So we are going to talk about two special functions here actually Which is related to integral and sometimes they are studied under improper integrals Okay, so these functions which I'm going to talk about they are called the gamma function and the beta function Gamma function and the beta function these functions are sometimes studied Along with improper integrals because they can be expressed as improper integrals now many people ask me What is an improper integral? It's just a name given when your limit of integration is Either becoming infinity or minus infinity so improper integrals are of this nature 0 to infinity Okay, or you can say you know minus infinity to infinity Because these are examples of improper integrals How many many question arise? How do we evaluate it? So it's not a difficult thing you evaluate these integrals by assuming that the upper limit is t and then take a limit of t tending to infinity That's not a difficult thing. So you do not need a special discussion on it We will understand most of the stuff when we are trying to talk about these two functions now What are these beta functions and gamma functions? Let me start with a gamma function So gamma function is basically defined as These are actually very helpful in some cases and many a times indirect questions can be framed on this in Jay See Jay syllabus doesn't specify beta gamma as such But what happens is when the question paper is set for advanced level the the Jay professors can put this as a Comprehension type or they can indirectly ask you because this is not beyond our basic understanding Right the the coverage of this topic is well within our basic properties that we have understood In our past in our previous classes, so it is not beyond those understanding So what's a gamma function? Gamma function is basically defined as an improper integral 0 to infinity e to the power minus x x to the power n minus 1 dx where and Here is a positive quantity This is basically written as this symbol gamma n Don't confuse it with the symbol. This symbol is factorial in Okay, there are two different symbols. This is a Bracket kind of a term upward. This is bracket downwards. Okay, just like your Ceiling function and floor function. So this is like your floor function. This is like a ceiling function half bracket Okay, anyways So this is how a gamma n is defined Okay, this integral is also called sometimes Oilers integral of the second kinds This is also called as the oilers integral of The second time Okay, I'll talk about oilers integral of the first kinds also, which is basically coming from your beta function Now these things will not be there in many of your J books, of course, it is there to a certain extent in Arihant But they're also they have not gone into details of it So I'll go into some details of it because if a question on this comes, I don't want you to be struggling with these Now a couple of things we will learn about the gamma function The first thing that I would like you All I mean I'm giving this as a question to you rather than as a direct property to you What is gamma 1 So let me name it as properties of gamma function. What is gamma 1? Can you all evaluate it gamma 1? So you'll say simple gamma 1 will be 0 to infinity e to the power minus x x to the power 1 minus 1 dx Right, so it is basically you are integrating e to the power minus x from 0 to infinity Which is nothing but minus e to the power minus x 0 to infinity When you put an infinity remember, you will get a 0 because e to the power minus infinity tends to 0 Basically when we evaluate this we do it in a very refined way. So we do it like this e to the power minus t then we take the limit on t tending to infinity Okay, minus of minus e to the power 0. So it's actually 0 plus 1. That's 1 so gamma 1 Gamma 1 is 1 Okay, please make a note of this Okay, what is gamma 2 what is gamma 2? Okay in the interest of time. I'll only do it as this Thing will take a lot of time. So e to the power minus x gamma 2 will be x to the power 2 minus 1 2 minus 1 is this Okay, can we apply integration by parts on this treating this as my first function treating this as my second function Okay, so integration by parts will give you x e to the power minus x with a minus sign 0 to infinity minus 0 to infinity Derivative of this will give you e to the 1 into e to the power minus x divided by minus 1. So that will make it a plus or 1 Okay, now this guy will anyways become a 0. Can I say that?  This is nothing but the previous result that we had derived. It is actually One again. So gamma 2 is also 1 Okay now gamma 3 if you Evaluate I'll not do the evaluation process for you. Gamma 3 will come out to be 2 Gamma 4 will come out to be 6 Gamma 5 will come out to be 24. Can you see a pattern in this? The pattern here is gamma n plus 1 is Actually n factorial but provided your n is a natural number. It doesn't work all the time Now in order to realize this we'll do a reduction formula We'll do a reduction formula I Would like you to derive this I would like you to derive this that Gamma n plus 1 is n gamma n Everybody please derive this. It's a very simple way to prove it.  Are you sure ma? So very simple if you see your left-hand side Left-hand side is gamma n plus 1 n plus 1 is 0 to infinity e to the power minus x into x to the power n plus 1 minus 1 Correct, which is actually x to the power n only Okay Let's apply integration by parts on this treating this as a you treating this as a V a Guys do let me know if I'm going fast. Okay Sometimes I don't realize I'm going fast. So please let me know Okay. Yeah, so x to the power n integration of this will be minus e to the power minus x 0 to infinity Minus Derivative of this will be n x to the power n minus 1 e to the power minus x with a plus sign here DX 0 to infinity you can clearly see that this day is going to be zero no doubt about it because Infinity will make e to the power minus x 0 and 0 will make x to the power n as 0 So either case you get a 0 and this will give you n times 0 to infinity e to the power negative x x to the power n minus 1 DX which is actually actually actually gamma n Gamma n so this is how you get n gamma n, which is your right hand side Okay, now because of this only we were seeing this property happening So if I asked you what is gamma 1 you could actually use the formula Gamma 1 or you can say gamma n plus 1 if you compare it your n plus 1 is actually 0 Okay, so gamma n gamma 1 will actually become what is gamma gamma 1 One it is because it will become what does it become? Zero gamma zero And how much did you get for gamma 1? Not one one Yeah, what is gamma gamma 1? Will be one gamma One one second Did we oh Sorry, sorry, sorry. Yeah, so now so zero gamma zero should be zero, right? Did we I'm sorry Sorry, sorry, we can only come from gamma 2. Yeah. Yeah. Sorry because if you have gamma 1 then what will happen? it only be Gamma gamma 1 basically is integral zero to infinity e to the power minus x x to the power n minus 1 right Yeah, so n n cannot be put as a zero because this guy will not be defined for a zero correct Gamma zero is not defined gamma zero is infinity. So we can only start from we can only start from two in this case Yeah Yeah, so gamma 2 will be 1 gamma 1 and 1 gamma 1 will be 1 gamma 3 will be Gamma 3 will be 2 gamma 2 Yeah, and you know who's already 1 so 2 into 1 that is actually 2 factorial Gamma 4 will be what gamma 4 will be 3 gamma 3 So 3 and gamma 3 is actually 2 factorial. So 3 into 2 factorial will be 3 factorial So what is happening here is that if your n becomes a Natural number then gamma n plus 1 will follow n factorial. That is what we have actually We are seeing from here. So if you go till gamma n plus 1 and n being a natural number It will become n factorial Is that fine by the way, I'll also add to this fact that gamma zero is infinity Gamma zero is infinity. Is that fine any questions here? Okay, now very important expression over here, which I will derive is gamma half Gamma half is given by under root of 5 a very important result which you will be using in solving many questions So if you know this result, it will make your life very very easy So, how do you first of all prove that gamma half is under root of 5? Now again, this prove you will not find in your j textbook. So listen to this very very carefully So what is gamma half? Gamma half as per the definition of gamma function is e to the power minus x x to the power n minus 1 n minus 1 will give you a minus half correct Yes or no? Now put x as Put x as y square That means dx is 2 y dy So this will become zero to infinity e to the power minus y square x to the power minus half will be what? x to the power minus half will be 1 by y and you have 2 y dy over here So y and y will get cancelled off giving you integral from Zero to infinity e to the power minus y square y dy Oh, sorry Yeah, sorry Okay Now what do we do here is? We make another function in the different variables. So gamma half. I can also write it as two Zero to infinity e to the power minus x square dx fine Now what do we do is we multiply both the gamma halves will multiply this gamma half and this gamma half So it will become gamma half square and this will actually become four times zero to infinity e to the power minus x square dx into Zero to infinity e to the power minus y square dy Okay Listen to this it basically gets converted to a double integral which we can write integral zero to infinity Again zero to infinity e to the power negative x square plus y square dx dy So how did that work? Okay, sorry So how did you like put in integral then integral? See basically what is happening is when you are integrating one of them you are keeping the other as a constant Because both are into getting integrated under different Differentials one is getting integrated with respect to x other is getting integrated with respect to y Correct. So when you write like this you first integrate with respect to x keeping y constant and then you integrate the y thing Both will give you the same result Okay Now there is something very interesting here Conversion of this double integral into polar coordinates. So right now you are seeing Cartesian coordinate will convert it to polar coordinates Now, what is polar coordinates? We all know in polar coordinates. We express things in terms of r and theta so let's say if you have R at an angle theta, okay, and you just change it to let's say dr dr Length, okay, and there's a small change in d theta. So what will happen? It'll create our DX dy area over here. This will be your DX dy area over here Now the same area DX dy. Let's say this is your Okay DX dy area. So what I'm going to do is I'm going to write the same DX dy area as what is this length? This length is rd theta and this length is dr So the same expression DX dy can actually be written as r dr d theta Now you learn later on that this conversion actually comes from the concept of Jacobians Okay, so Jacobian is an expression which converts which converts The transform coordinates differential in terms of the original Differentials of the variable for example, let's say I'm not using I'm not using r and theta So my this term would actually be dr d theta and this Jacobian is basically an expression which is given by Doe X by do you? Doe X by do we By do why by do you? Doe Y by do we Now you don't have to know this actually. I'm just giving you a brief idea how it works So let's say you want to convert DX dy. Let's say this is your DX dy In terms of dr d theta, you'll realize that your Jacobian will come out to be our How very simple let's say let's say Your X is our cos theta and why is our sine theta, correct? What is do X by do are do X by do are that means what is the derivative of this with respect to R? So you'll say cos theta Okay, what is do X by do theta? This will be minus our sine theta Similarly, do why by door do why by door will be sine theta and do why by do we means derivative of this with respect to theta Which is our cos theta if you evaluate it you'll get our cos square theta plus our sine square theta, which is actually your R So what it says is that you can convert DX dy into our dr theta Now this is something which is very extra.  How does DX dy become our dr So one way is to look through Jacobian angle, which is Future concept for you other is by directly seeing this area So any DX dy area can be obtained by doing our d theta multiplied with dr Now, how does it help to solve this question? So if you see this integral If you convert everything To our square so this will become negative R square because you're converting all the things to polar coordinates DX dy will become our dr d theta Now in order to come cover all the space from zero to infinity You'd agree with me that R will go from zero to infinity and theta will go from zero to Zero to Pi by two see What are you doing here? You're trying to cover all the area in X from zero to infinity, right? And in why also from zero to infinity? So if I have to play the same role in terms of R R will go from zero to infinity and angle will go from So you're you're trying to cover this entire area, right? So your angle will go from zero to pi by two only Correct, so this gets converted to a double integral in terms of R and theta So it'll be from zero to infinity and theta will be from zero to pi by two Now integrate this as two separate integrals keeping your theta separately and Keeping your R term separately Okay, now this can be easily be integrated if you take your R square as a K So R dr will become K by 2 DK Okay, so it becomes zero to infinity e to the power minus K RDR Sorry my bad Not K DK just DK by two, okay, so this will become DK by two DK by two and Here also it is just four into pi by two Correct now. What is the integral of e to the power minus K from zero to infinity? We have just seen it is actually gamma one gamma one is one So pi pi will go off. Sorry for four will go off and pi into one will be left off In other words, you have got gamma one Gamma half square as pi so gamma halves Will be under root of pi Just remember this is that you don't have to go into details of how they are coming because It goes into the concept of double integrals which you have not been exposed to before But you'll definitely thank me when you're going in your first year of engineering and your professor is teaching you all these things Then you'll remember that oh, I guess I had taught this in our class So like can you find gamma of a negative number as well? Gamma can only be found out for a quantity greater than zero Okay, now other numbers are very difficult to find out for example if you say what is gamma three by four? Okay, that would be slightly difficult to find out We always deal in multiples of half gamma half gamma one gamma three by two Because that is what we need for our problem solving You'll realize soon that we don't require any other gamma than these values either you need gamma of natural numbers or you need gamma of You know multiples of half so gamma half would be required gamma three by two five by two seven by two Those will be required so beyond that I would not like to go But however, I'll give you certain properties which probably some of them will try to prove some of them I'll leave up to you to prove The next thing that I would like you to understand is this function can I go to the next page? So I told you in the beginning of the class. You'll find things slightly difficult in this session but the Despite is yes. Yes, okay. Next thing is next property is Gamma Gamma of any quantity. Let's say Let me write it as gamma z. Okay, and gamma one minus z Is basically given by Pi by a sine n pi Okay, n should not be an integer here. In fact n should belong to zero to one here. This formula is called the Oilers Reflection formula now, how does this come out of this formula actually come? Let's talk about the formula See we all know we all know that E to the power I'll just try to prove it and again the proof is not required only the result is supposed to be known Let's say we all know that e to the power. I can write any function. Let's say minus t Okay, can I write it as a limit which is limit n tending to infinity? One minus t by n n to the power of n. You all know your limits, right? One to the power infinity form Can I obtain e to the power minus t from this limit? Yeah Everybody's convinced Okay, so if I say gamma z as for the definition of gamma z Gamma z is what? zero to infinity t to the power z minus one Or you can say let me write it like this e to the power minus t into t to the power n minus one Sorry z minus one since I've written my n as a z I can write it like this Okay Now what is e to the power t I can replace this with e to the power t So I can write this like this limit and tending to infinity Zero to n This term I'm replacing it with this term Okay, so I can write it as one minus t by n to the power n Okay, and e to the power sorry and t to the power z minus one. I'll copy as it is Okay, anybody has any doubt so far. No problem with this okay now integrate this by parts integrate this by parts Now for integrating this by part Take this as your v take this as your u Okay, so gamma z can be written as limit and tending to infinity Yeah, so it is you Integration of this will be t to the power z by z Okay Integral from zero to n minus Derivative of this term will be n Into One minus t to the power n to the power n minus one and also you'll have a minus one man because you'll be applying Chain rule over here. So this will become plus and this will become n and Integral of this will be t to the power z by z Dt from zero to n. So as a result, what will happen when you put any you'll get everything as zero Okay, so what will happen to this result? So this result will give you one by z Integral zero to n one t to the power z One minus t by n to the power n minus one Okay, now we'll apply induction over here We'll apply induction over here Can I say if I do this operation n times What is going to be the ultimate result that you're going to see? if you integrate it by parts and times and Times you will figure out that your gamma z will ultimately boil down to limit and Tending to infinity this one by z. I will write it as n by n z Then you'll have n minus one by n z plus one Then you'll have n minus two by n z plus two this you can try it out at your end also in the interest of time I will just write down the final result and Finally, you will be left with this times integration zero to n T to the power z plus n minus one dt Okay, so I've done it once but you have to apply it n times To check that this is the gamma n expression gamma z expression that you will be getting okay this term This term if you see it is actually n factorial on the top and You have n n n occurring n times so n to the power n will come correct and You will have a product of You will have a product of Product of one by z plus k Okay times integral of this Now what is the integral of this? Let me write k from zero to zero to To n right or n minus one Yeah, zero to n minus what are the integral of this Integral of this is going to be I'm just doing it separately T to the power z plus n minus one plus one Okay, zero to n So it will become n to the power z plus n by z plus n am I right? Okay, so what we can do here We can write this as integral Now one by z plus k if you are multiplying it from zero to n minus one and there is one more term coming up So can I say just take this into the purview of this term and make it n over here? Can I say so any questions here up till now? Okay Now let us cancel out n to the power n from this n to the power n over here So I will have n factorial or you can take out n to the power z also here Okay, n to the power z And n to the power n will get cancelled off. So it'll be product of K equal to zero to n one by z plus k. Okay So far so good. No problem with this Sorry So what do you have there? Oh, yeah, in fact And factorial what I'll do is just correct me if I'm wrong. Can I include this as a k over here? Yes Okay Now I can write this as n to the power z by z product of Z plus k from k equal to one to n now see what I did as zero thing I put it over here and took it out because when you put a zero you'll get What do you get you'll get the you know Sorry sorry No, is this step from here to where is it any any mistake is there just tell me Oh, sorry, this this will be this will remove a zero. Yeah, sorry This will be one any any mistake from here to where just tell me I'll rewrite it again rewrite it again once again This will be k equal to zero Okay Now what I'm doing is I am breaking this up as n to the power z into n to the power n So n to the power n into the power n will go off n to the power z will come out okay, and we'll have and factorial also product from zero to n one by z plus k So far so good no problem with that. Yes, sir. Okay. Now what I'm doing is When you put a zero here it'll come it'll become One by z and Can I say the other terms would be starting from one to n? Now it is proper Richard Yes, yeah So now I'm going to take this to the next page because I'm running out of space here So gamma z I can write it as n to the power z by z product of product of K by z plus k Okay, this I can write Don't forget limit and tending to infinity. So this I can write limit and tending to infinity Z to the power sorry n to the power z by z product from One to n one by one plus z by k. Okay Now if you look at if you look at The property that we had discussed gamma n plus one Gamma n plus one is n gamma n right? Right. We have seen this property correct so If I replace my n with if I replace my n with minus z Can I say gamma one minus z is minus z Gamma's Minus it correct. Yes. Hello in other words. Can I say? Can I see Gamma z into gamma one minus z is Going to be minus z gamma z gamma minus z So I multiplied both sides with gamma z. Yes. Okay. Now. I already know gamma z expression is this This is your gamma z expression Right, and I need this expression So, can I say gamma z gamma one minus z is minus z gamma z? I'll write this term which is Which is n to the power z by z product of Product of k equal to one till n one by one plus z by k times times Gamma minus z gamma minus z will be n to the power minus z by minus z K equal to one till n one by one minus z by k Can I write it like this? Okay, both of them limit and tending to infinity correct So ultimately what do you see? Ultimately I see is that This minus and this minus will be taken care of one of the z's will go off This and this will neutralize each other. So it'll become product of one by one minus z square by k square So it'll actually become this Okay, so this will become your gamma z by into gamma one minus it now now If you look at sine pi x expansion or sine pi z expansion By Maclaurin series What does this expansion say? It is pi z Minus pi z cube by three factorial and so on correct Am I right? if you multiply with a pie z or Let's say if you divide by a pie z you realize that You will end up getting one minus One minus Pi z square by three factorial Plus pi z to the power four by five factorial and so on Okay But this term is what this term basically you can write it as One minus but from K equal to one till infinity. Can I write it like this? Can I write it like this? Actually, this comes from your oilers formula oilers formula, which probably I mean this is this requires a bit of more Research because it comes from the oilers formula That is pi square by six is equal to one plus one by two square one by three square and so on I don't know whether you have learned this This is pi square by six Are you aware of this formula anybody? Okay, I'll leave up to you to figure this out now just quickly proving it again I it's not required to know the exact proof for this So if you see this term is occurring in the denominator over here So if you take a reciprocal of it, so it becomes pi z by sine pi z pi z equal to Big pi k equal to one to infinity one by one minus z square by k square So if you bring this z down Okay, it actually becomes the same expression that we were actually looking for this expression. So this is Nothing, but one by z limit and tending to infinity big pi k equal to one to n one by one by z square by k square, which is actually your Gamma z gamma one minus it Okay, now, this is something which I would like you to prove out I've given a hint how to do it. I think in past also. I've discussed this with you. So please prove this and Let me know Okay, if you are stuck, I'll derive it and send it to you on the group. So this is called the gamma Sorry, it is called the Oilers Reflection Formula Okay, no need to go into much detail of this. It's just if a question like this comes you should be ready to handle it Now a very important property that we would discuss right now is Gamma m plus one by two Gamma n plus one by two By two gamma m plus n plus two by two is Integral zero to pi by two sine to the power mx cos to the power nx Okay, m and n should be positive over here. Now how to prove this I'll prove this when I talk about beta function as of now. Let us note it down. We'll prove it when we are doing beta function, okay Have you all noted it down everybody? Let us take some questions on whatever we have done so far Else these concepts will not stick for long. Let us evaluate Let us evaluate a very simple question integration from zero to one Log one by x to the power of n minus one. How do you evaluate this? Or let me make some values over here. Let's let's let's say I make a value five over here Log one by x whole raise to the power five integral from zero to one Okay, because very good only because could solve it Okay, Richard. Okay, so what do we do in this case? What is the approach in this case? We take our log of one by x to be t Okay, that means one by x is e to the power t. That means x is e to the power negative t So DX will become negative e to the power negative t. So what will happen to this integral? log of one by x is like t to the power five DX is like e to the power Negative t with a negative sign dt Now what happens to the limit of integration limit of integration will actually become from minus infinity to zero check it out because when you put a zero It becomes infinity when you put a one it becomes log one log one is zero right now If you switch the limit position, you know that you can Observe the negative sign Now just try to compare this try to compare this with it Zero to infinity e to the power minus x x to the power n minus one DX You can clearly see that your n minus one is actually your five So this is an expression for gamma six Now people who do not know this formula what they would do they would sit and do integration by parts five times or They will have to use the reduction formula Okay, so quickly as a person who knows gamma function You can write five factorial as the answer which is 120 are you getting this point? So it's very easy to solve this question if you are aware of the gamma function any questions here Meanwhile keep thinking about the proof for the previous problem. I'll discuss it towards the end of the today's class Next is your beta function. This beta function beta function is basically a Function which takes in two parameters m and n and it is defined as zero to one X to the power m minus one one minus x to the power n minus one DX Okay, m and n have to be positive quantities now Right now I've expressed this function as a proper integral, but this can also be expressed as an improper integral So the first thing that I would like you to show me is that number one? beta function Can also be written as Can also be written as zero to infinity x to the power m minus one by one plus x to the power m plus n Okay, and of course it can also be written as zero to infinity X to the power n minus one by one plus x to the power m plus one n First prove this any one of them you prove Let's say you prove this one So this could be a question in itself Jay can ask you that this integral is equal to which of the following integrals and one of the options or you can have multiple options sitting like this okay Let me take this up First of all proving this and this equal is very easy because you can always show that Beta mn is same as beta nm Okay, how do you show that by use of King's property? So by King's property You can write one of the integral let's say zero to one x to the power m minus one one minus x to the power n minus one dx as Zero to one one minus x to the power m minus one one minus one minus x Whole raise to the power of n minus one dx So it is actually zero to one x to the power n minus one one minus x to the power m minus one dx That means your m and n can interchange positions If your m and n can interchange positions, that means your beta nm is same as beta m Okay, so same thing here also have done if you change your m and n positions this will convert to this Okay, now, how do I get this expression per se? Let us find that out. So what I'll do number one what I'll do is I will integrate This by the use of certain substitution. So what I will do I will substitute x as One by a one plus y let's say Okay, I'll substitute x as one by one plus y or in short. I'm substituting y as one by x minus one Okay, so dx will become minus one by one plus y whole square dy Correct So if you put it over here What happens to the limit of the integration when you put x as zero y will become infinity When you put x as one y will become a zero. So as you can see I have converted from a proper to improper now Let's simplify the other terms. This will become one by one plus y to the power m minus one Now what about one minus x One minus x would be One minus one by one plus y which is actually y by one plus y whole this to the power n minus one and dx would be replaced with negative one by one plus y square dy Correct me if I'm wrong Negative sign could be observed by flipping your limits of integration So make it zero to infinity instead of infinity to zero numerator will have y to the power n minus one While in the denominator, you can check you will have one plus y to the power m plus n And since there is nothing in the name of the integral Sorry, nothing in the name of the variable. I can put this back to this Okay, which is anyways same as zero to infinity x to the power m minus one by one plus x to the power m plus n So this is something which was beyond not beyond you Okay, you could have done this problem had you applied your Substitution that's why we did that exercise on certain Integral can be expressed as another integral when you apply a substitution to that question So it belongs to those kind of questions Now in your Arihant book that you you must all be having many of you are following Arihant book They have not derived this they have not proved it. So let's move on to the next property Done copied. Anybody wants me to scroll it somewhere. Can I go to the next page? Next important property is the one which connects the two functions beta function and gamma function Now there was a question which I left I know unfinished which was I think in the previous to previous slide. I'll just show you that question Yeah, this part this question Yeah, I told you when I discuss beta function with you I'll discuss this part So now time has come that will prove this also, but before we prove it We need to prove This guy. Yeah We need to prove this. How will you prove this? Okay, let's let's go back to our basic definition of a beta function so beta function is Integration from zero to one x to the power m minus one one minus x to the power n minus one This is a very basic definition of beta function correct correct no Yes, any doubt about it? Okay. Now what I'm going to do here is I'm going to replace my x with sine Square theta Okay, when I replace my x with sine square theta, what do I get? First of all, your DX will become to sine theta cos theta d theta, right? That means this entire integral becomes zero to pi by two This will become sine to the power two n minus two theta d theta One minus sine square will become cos square. I'm sorry One minus sine square will become a cos square. So it will become cos to the power two n minus two theta d theta and This term Why am I writing d theta? Yeah, this term will become to sine theta cos theta d theta correct in other words your beta mn I Can write it as two times integral zero to pi by two Sine to the power two m minus one theta cos to the power Two n minus one theta so far so good. No problem with this. Anybody has any problem with this? Okay, so I would like you to note this down because this is yet another convergent of beta function So very important. Please note this down. Okay now Now let me go back To gum. Oh, sorry. Sorry. One second. Well, yes, sir. Guys. How are you finding this? difficult Yeah or extremely difficult Okay, don't worry. I'll tell you which points to focus on which points not to focus on because Normally, I see the batch and prove it. Okay, if I'm I mean, I don't know to say this on recording But when I when I see a batch is not able to understand a simple thing Then I'll never prove these things because it'll end up, you know, blowing up their mind But you are a batch and you can take this as a compliment who reciprocate very well to my questions So that's why I'm proving it in front of you If you see my previous videos, I don't prove it in all the batches Normally, I see the batch and prove it Okay, so I'm proving it because I know you can grasp it So let's say I start with gamma M definition. Gamma and definition is what zero to infinity e to the power minus t t to the power m minus 1 dt Let's put t as x squared That means your dt is 2x dx Okay, so can I write gamma M as 0 to infinity e to the power minus x square T to the power will become x to the power 2m minus 2 and dt will become I'm sorry Yeah, sorry, sorry dt will become 2x dx So it'll become 2 times 0 to infinity e to the power negative x square into x to the power 2m minus 1 dx Correct. Okay In a similar way, I Can say gamma n could be written as two times I'll just change the name of the variable to y. Let me multiply gamma m with gamma n Okay, now the same story that I discussed while proving gamma half is equal to root pi will be applicable So it is four times zero to infinity zero to infinity e to the power minus x square plus y square x to the power 2m minus 1 y to the power 2n minus 1 dx dy correct Yes or no Now let us use polar coordinates. Let us use polar coordinates. Let's substitute x as r cos theta and y as r sine theta So gamma m gamma n would be four times Correct me if I'm wrong It'll become zero to pi by 2 zero to infinity e to the power minus r square This term will become r to the power of How much? Uh Sum of these two right so 2m plus 2n minus 2 into cos theta to the power 2m minus 1 sine theta to the power 2n minus 1 Into dx dy dx dy was r dr d theta remember the Jacobian was r correct now one of these r's You can drop and change this power over here Okay, so I'll write once again. So this you can write it as Again, I'm splitting it up two times zero to pi by two all the Term containing theta. I will bring it to one side Okay d theta and all the terms containing r I'll bring it to one side agreed This is gamma m gamma m agreed yes or no Now A couple of minutes back A couple of minutes back. I discussed that this term was actually what? Check your notes. This term was actually what? Beta mn correct. This is gamma m gamma n And what is this term? This is actually gamma m plus n Why because you can substitute your r square Back as a x. Let's say if you put r square back as an x correct Then dx or you can say dr is dx by 2r correct, so this will become two zero to infinity Remember x will also go from zero to infinity e to the power minus x This will become x to the power m plus n minus 1 am I right Why because if I put a 1 by r over here See dx is what dx is I'll go back to the term again So it is r to the power 2 m plus n minus 1 dx is what? Or dr is what dr is dx by 2r So this 2 will kill this 2 And this r will increase the power of this to minus 2 Right minus 2 means you can write it as twice of m plus n minus 1 Which is actually x raised to the power this guy is x So this is x to the power m plus n minus 1 Okay, so as rahul rightly said this is actually gamma m plus n And therefore you can prove the relationship between beta function and gamma function, which is very very important Please make a note of this So this is a relation which connects the two functions gamma functions and beta functions very important got the point now The proof which I left off in the previous slide time has come now to prove it So let's go back to that So first, please make a note of this you'll not find these proofs in your Any of the j books? Let me tell you done copied Copied everybody.  I used to call it walley when I was preparing Walley's formula Yes, the proof is very easy. We had actually we have actually proved it See the proof comes from the fact that this term was actually written as two times zero to pi by two Just check your notes We had written like this, right Correct a little while ago while I was proving the relationship between beta and gamma function. I wrote this correct Yes, and this term Now I have proved that this term can also be written as gamma m gamma n by gamma n plus n Correct. So what I do now I'll make a very smart move I will write this guy as some capital or you can write it as m dash Okay This guy 2 n minus 1 you write it as n dash Correct in other words your m will become m dash plus 1 by 2 And your n will become n dash plus 1 by 2 you do one thing you replace In terms of m dash n dash everywhere. So this guy will become m dash plus 1 by 2 This guy will become n dash plus 1 by 2 Denominator will become the sum of these the sum of these two which is actually m plus m dash plus n dash plus 2 by 2 On this side, you have a 2 that you bring it down over here Okay, and this becomes zero to pi by 2 sign 2 m minus 1 was m dash correct And 2 n minus 1 was n dash Now change your name of m dash n dash back to m and n because name doesn't matter Okay, so change m dash back to m n dash back to n So you will end up getting zero to pi by 2 Sign to the power m theta cos to the power n theta d theta Is equal to Gamma m plus 1 by 2 gamma n plus 1 by 2 By twice of the sum of these two which is m plus n plus 2 by 2 Very important formula very very important formula. Please remember this result When m and n are natural number, this is basically called as the the wall is formula Now Having said so What are the things which we really need to take home from this 1 hour 15 minutes of You know hard work that we did Few things only you need to take care. So I'll write a summary here just to you know give you a final sigh of relief Oh my god, so you have taught so many things which some of them went above our head So what actually you need to know from this entire exercise? Number one, you need to know what is the definition of beta function and gamma function number one You need to know the definition which is 0 to infinity e to the power minus x x to the power n minus 1 dx you need to know beta 0 to 1 x to the power n minus 1 1 minus x to the power n minus 1 Okay, this is number one definition by the way You should also be knowing 0 to 1 x to the power n minus 1 1 minus x to the power m minus 1 that means beta and Beta m comma n is same as beta n comma m No difference between changing the position and you should also know its Improper form that is x to the power m minus 1 1 plus x to the power m plus n Okay, of course needless to say it's 0 to infinity x to the power n minus 1 1 plus x to the power m plus n also Okay, this is number one thing you need to know number two that you need to know is that The recursive relation that gamma follows gamma n plus 1 is gamma n gamma n and in particular And in particular you can say If your n is a natural number then Gamma n plus 1 is n factorial Okay, second thing you need to learn like even for Uh multiples of half We can do a similar thing and just keep going until Yeah, that's what we're going to see in some of our problems coming up param. Okay third thing you need to know Is gamma half is root pi Okay, approximately 1.77 Okay Fourth thing you need to know is that beta m n And gamma function are related by this relation Okay Fifth thing you need to know in fact, I should have written it over here also here only But I'll write it again in a separate form fifth thing you need to know Is integral 0 to pi by 2 sine to the power m theta cos to the power n theta Okay, see the expressions are very symmetric So even if you stop the position of m and n nothing will happen Okay, it is given as gamma m plus 1 by 2 Gamma n plus 1 by 2 by 2 gamma m plus n plus 2 by 2 that's it You need to know nothing else beyond this Okay, no proof is required for you to know only these five things if it is clear in your mind Life is set I mean life is set with respect to beta and gamma function Now what sir you settled our life only with five formulas Sir we can have gamma half No, no the oil is reflection formula not required Which by the way, I wanted you to go back and Reflect a bit on the formula which I wrote for you. I think Nobody bothered to Prove that formula see what what did I write actually I wrote something How do you prove? How do you prove this? How do you prove this? Okay, this is actually called the basal's formula Basal is a place in Switzerland and Euler came from that place. Okay So it is a basal formula. How do you prove this the The agenda behind this prove will actually tell you the proof for the thing that I had written in the In a previous to previous. I think long long ago. I wrote that. Okay. So how does it come actually see we all know sine x expansion correct What is sine x expansion? x minus x cube by three factorial plus x to the power five by five factorial and so on correct This is something which we get from our maclaurin series Maclaurin series Okay, now If I If I do sine x function, right Can I say is sine x function or I can say let's say sine x by x function. Can I say this function Has a root of pi Two pi three pi four pi and so on minus pi minus two pi etc Yes, okay So can I write such a term as Can I write such a term as One minus x by pi One plus x by pi One minus x by two pi One plus x by two pi and so on One minus x by three pi One plus x by three pi and so on Okay, so I can keep on writing this. So basically I'm trying to factorize this because I'm I'm trying to factorize this because I know this particular expression will have roots of Pi two pi three pi etc Minus pi minus two pi minus three pi etc. So do you have any objection in me writing it like this? Okay, so then can you also do that first sine x? Yes, yes, yes Of course, we can do that This basically expression is was given by V stress. That is why it is also called V stress factorization theorem Okay, so V stress was the first one the spelling is w e i e r Okay V stress inequality also people who would have prepared for rmo and all would have prepared Okay, so V stress suggested that sine x by x could be actually written like this Okay, so if you can write sine x by x Like this you can also write sine x like this then how does that make sense? sorry So you said like we can also write We can also write it like this. Okay. Now. That's what I'm trying to compare So very very well point brought out because that's what I'm going to do So if you divide this by x you'll get one minus x square by three factorial plus x to the power four by four Pi factorial and so on and this is also equal to this by the way if you club this up it becomes one minus x square by pi square Then one minus x square by four pi square One minus x square by nine pi square and so on Now what did Euler did what did Euler do? He compared the coefficients of x square on both the sides He compared the coefficient of x square on both the sides by the way Our proof is over here itself. Remember I had written this big pi x square by k square pi square Sorry k pi square Did I forget that k pi square there just make a change there? Okay, so What Euler did was he went a step ahead then waste us and he compared the coefficient of x square on both the sides So here he got one by six and here if you check his coefficient will actually be Minus one by pi square minus four by pi square minus one by nine pi square and so on till infinity Correct Because to get a minus x square you need to choose one by pi square from here minus one by pi square and one one one from all the terms, isn't it? Same for other factors also So what he did was he multiplied both sides with a pi square minus pi square So he ended up getting pi square by six is one plus one by four plus one by nine all the way till infinity And that is what makes the basal formula So pi square by six is one plus one by two square one by three square and so on So I had one doubt sir. I think it was trying to say the same thing So wasn't the sine x expression the same as sine x by x expression through that logic That pi and 2 pi and all our roots So if we don't put the x in the denominator, it still works like the same logic No, how is this expression same as this expression? So the logic was that pi and 2 pi are roots right and then there's nothing in the x in the denominator Like that doesn't feature it right? No, both will have the same set of roots, but that doesn't mean the expression will be the same So then like why can we make sine x by x in this form but not sine x Yes, so because Both of them share the same roots pi 2 pi minus pi minus 2 pi, but Yeah, if you if you divide this by x it should be equal to this that is what I'm trying to say That's how we got the formula No, sir, but like so when you said that sine x by x should look like this the logic was that We wanted it to have all those factors Yes, so then why can't we use the same logic for sine x and then say that sine x is also equal to one minus x by pi into One plus x by pi like that Because even sine x is only zero at those same points Okay, now from where did Vistras get this expression? He got this expression from the fact that as limit extends to zero your answer should actually give you a one So that will happen when you take the limit extending to zero over here Does sine x do the same thing? Sine x as extends to zero in that case will give you a one in that case which is wrong because sine x extending to zero should give you a zero Okay, so this is just an expression which Vistras proved but I'm not Going into the proof of this, but I'm trying to justify why He actually chose this expression and why it is actually correct also it works And why your formula will not work? Yes Okay Good, so I think we proved the Euler's reflection formula also But this is just for your extra Gyan here. I mean nobody is going to pester you on that Let's take few questions, but on the other side of the break. I think we all need a break Okay, welcome back after the break. I hope everybody is back Okay, so last Before the session before the break session we discussed the concept of beta gamma function now time to apply it Let's see how can we solve functions very how can we solve integrals? Very quickly by the use of beta gamma functions Let's say I want to solve this integral So I'm not denying that you cannot solve it by Any other way other than beta gamma function? I'm not saying that you can use any other way to solve it Okay, you can use your substitution of x square as or x as sine theta cos theta whatever you can do it But if you know your beta gamma function, your life will be very easy. You can you can write these results very quickly Okay, so first I leave up to you to try this out Actually, I gave you a hint. Yeah, you can try that out try that out Richard. No problem Okay, well, I should solve this question so that you can get an idea how to approach see Richard suggested substituting x as sine theta Okay, fine. Fine. I'll I'll wait for some time Okay So dx is equal to cos theta d theta, correct So your limit of integration will become zero to pi by two Uh, what does this become this becomes sine to the power six theta This will become cos theta to the power of half Correct. Oh, sorry cos theta to the cos theta to the power Yeah, sorry cos theta to the power of two d theta Now if you see this term It would remind you of the integral zero to pi by two Sine to the power m theta cos to the power n theta d theta Which was expressed as a gamma function Like gamma m plus one by two gamma n plus one by two By two gamma m plus one m plus n plus two by two, correct. Now here The role of m is being played by six And n is two over here So as for the formula your answer will become seven by two gamma three by two by two times gamma five now Gamma seven by two is actually five by two gamma five by two which is three by two gamma three by two which is half Gamma half which is root pi So this term that is the first term you can directly write it like this, okay Gamma three by two is half gamma half which is half root pi The denominator is two into gamma five is four factorial which is 24 Okay, so the answer to this question is going to be Let me just drop this three and write an eight One two three four five Six seven eight two to the power eight two to the power eight is 256 answer is five pi by 256 That how quick it was Okay People who do not know this what they will do they will apply that I know When you have even powers on sine and cos Remember we did a complex number approach They will apply that thereby wasting a lot of time Thereby wasting a lot of time if you know this you can do it pretty quickly if you just know this formula So this is the important formula here that proved very helpful Let's take another one Try this one out quickly. You should give me within one minute. I should get the answer Sine to the power four x Cost to the power three x Done Okay, I've got responses for many of you. Oh my god different different answers. I'm getting okay So the answer for this is gamma five by two Gamma two By two gamma nine by two remember I'm using this formula the wall is formula here So m is four here. So four plus one by two will come that is five by two n is two here Sorry three here. So three plus one by two, which is four by two which is two By two and the sum of these two some of these two is nine by Two, okay Now gamma five by two is nothing but three by two gamma three by two which is half gamma half, which is this Gamma two is one By this will be two into seven by two five by two three by two into half root pi Up till this will get oh, sorry Up till this will get cancelled And your answer would be two by 35 Mahid got it. He got it Rich's thing got it very good. Why are those made a mistake? Yes, sir, sir, could you show that formula once again? I think I wrote one sign No Yeah, this formula. Yes Yes zero to infinity x to the power n plus one by one plus x dx Is pi by sine n pi n should not be or you can say n belongs to Zero to one. Yeah Using Euler's reflection formula, this says gamma n Gamma one minus n Is pi by sine n pi Beta m comma n we all know it's gamma m Gamma n By gamma m plus n right Take your m as n and take your n as one minus n Right, so it'll become beta m sorry beta n One minus n is equal to gamma n gamma one minus n and this will become gamma one Gamma one is known to be one. So this will go off So it will become just this okay now this term This term Is pi by sine pi n And this term which is on the left hand side If I write it as an improper integral it is x to the power n minus one by one plus x to the power n plus one minus n right As per the very basic definition of beta function So integral zero to infinity x to the power n minus one by one plus x dx Is pi by sine pi n n should be between zero to one Is that fine? Why is this called reflection self formula reflection power? Actually when this was derived by Euler he used contour integration for it And the contour that he got was the reflection of one of the other you can prove this by using your integration on complex domain Which is obtained by making contours For that you have to know your residue formula and all those things kushi's residue formula So I'm not going to details of that. It's it's a vast domain of complex integration Surely in your first year or second year of engineering you will learn it Okay, that's why the name reflection was given to it Okay, okay, let's do this one zero to infinity e to the power minus a square x square dx Is which of the following under root pi by 2a pi by 2a pi by under root 2a No tough I'll put the poll on for this Almost two minutes gone only two people have responded so far If you cannot see the poll you can put it on the chat box Okay, para mil check last 30 seconds Last 30 seconds, okay five four three two one go Okay, most of you have given the answer as a Okay, let's check So here we can go for a substitution So let us call a square x square as a t Okay, so your x square is 1 by a square times t that means dx first of all here x is under root t by a so dx will be dt by a square by 2x it's correct So when you make this a placement you end up getting minus t dx is dt By 2a square x is root t by a So this will kill away this one of the a's Now it becomes super easy question. It becomes 1 by 2a e to the power minus t t to the power minus half dt Compare this with n minus 1 so n will become half. So this is actually 1 by 2a gamma half Gamma half is root pi so root pi by 2a will be the answer option a is there So janta was correct option a was the right option Evaluate evaluate talk 4 to the power 5 times integral 0 to infinity x to the power 4 by 4 to the power x dx Yeah, I think small mistake happened, but I'm there I thought it was easy question Nobody wants to respond Okay, take your time That's correct here very good See 4 to the power x Right Can we write it as e to the power something? Can I write it as e to the power ln? Can I write it like this? Can I write 4 to the power x? Can I write 4 to the power x as e to the power x ln 4? Right? Yes So ln 4 to the power 5 0 to infinity x to the power 4 this 4 to the power minus x I can write e to the power minus x ln 4 Can I write it like this? Please correct me if there is any mistake here Now what I'll do is I'll take x ln 4 to be y That means x is y by ln 4 So dx and dy would be the same Sorry dx will be dy by ln 4 Okay So what are the benefit that I get by doing this? I get ln 4 to the power 5 Okay y will also go from 0 to infinity no doubt about it x to the power 4 will become y to the power 4 By ln 4 to the power 4 This will become e to the power minus y dx will become dy by ln 4 Now we can clearly see that ln 4 ln 4 to the power 4 will get cancelled with ln 4 to the power 5 Ultimately leaving you with e to the power minus y y to the power 4 dy which is actually gamma 5 Gamma 5 is 24 Is that fine? So only he got this correct Else everybody made a mistake very good here Next question Can I move on? Can I move on to the next question? Okay next question is Evaluate 0 to infinity dy by 1 plus y to the power 4 Very good Param Okay Anybody else who wants to respond Other than Param Param, how did you do this question? I've done the indefinite integral before Okay I've practiced it so many times that I skipped over most of the steps and just plugged in the Limits, okay anybody else who found a easier way to do it Partial fractions, how do you use partial fractions? Is it factorizable? Okay, let's say y to the power 4 is x Okay Can I say 4 y to the power 3 dy is equal to dx? Correct That means dy is one fourth dx By y to the power 3 what is y to the power 3 y to the power 3 x to the power 3 by 4 correct If I place it over here it becomes 0 to infinity one fourth One fourth x to the power minus 3 by 4 dx By 1 plus x Okay Can I write this term as 0 to infinity x to the power 1 by 4 Minus 1 Correct Compare this term With 0 to infinity x to the power n minus 1 by 1 plus x whose result was already known because of the Euler's Selection formula which was gamma n gamma n minus 1 And n has to be between 0 to 1 which is very much satisfied by this guy one fourth Okay, and this result is known as Pi by sine n pi So your answer will be one fourth pi by sine pi by 4 Correct that's nothing but 1 by 2 root 2 Is this what you got people who answered Yes So do you realize how easy the formula becomes the integrant integrant becomes because of the use of Euler's reflection formula See that that's where you will save time. I'm not teaching this for you know scaring you. I'm not teaching this for Future I'm teaching you so that if you know this you will if at all it is required you can save a lot of your time Where others would be busy applying your indefinite integrals You will be able to solve this within one minute Okay, based on the same I would like you to ask this question I think this you should be able to answer pretty quickly. You can try it out. Adwik. It should work. It should work None Correct. Yeah Oh, sorry, correct. Kirtan correct here good Again use the fact that beta m comma n is 0 to infinity x to the power of m minus 1 by 1 plus x to the power m plus n Okay, and it is also equal to Correct richer richer thing. It is also equal to this Okay Now if you treat your m as let's say 11 And n as let's say 19 What will you get from here? What will you get from the first one? The first one will give you 0 to infinity x to the power 10 by 1 plus x to the power 30 And the second one that is this integral will give you 0 to infinity x to the power 18 by 1 plus x to the power 30 Both are same because both correspond to both correspond to beta 11 comma 19 or beta 19 comma 11 both are same And since both are same their difference, which is actually the given integral will actually be 0 Is that fine? Okay So with this we end this beta gamma game