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Improper Integrals

  1. Sep 30, 2008 #1

    Calculate [tex]\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}x^{2}e^{-\frac{x^{2}}{2}}dx[/tex]

    Use the fact that [tex]\int^{\infty}_{-\infty}e^{-\frac{x^{2}}{2}}dx=\sqrt{2\pi}[/tex]

    2. Relevant equations

    I'm assuming that integration by parts is the best way to solve this.

    http://www.math.hmc.edu/calculus/tutorials/int_by_parts/" [Broken]

    3. The attempt at a solution

    I want to use integration by parts in order to solve this. I've attempted both combinations of u and v'. However, I'm not able to get an integral in any of my solutions that looks like the one above that is set equal to [tex]\sqrt{2\pi}[/tex].

    I have plugged the equation into Mathematica and the answer comes out to be 1. However, getting my work to back that up is proving to be difficult.

    Is integration by parts the right way to go about solving this one? Any help is greatly appreciated.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 30, 2008 #2


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    Homework Helper

    The integrand is the density of the standard normal distribution, which is why Mathematica gave you the answer it did.

    You should try integration by parts; with a judicious choice of [tex] dv [/tex] you will find that the

    \int \, dv

    portion (to calculate [tex] v [/tex]) is easily done. After the step, and you have

    \int u \dv = uv - \int v \, du

    you will see the reason for the hint.
  4. Sep 30, 2008 #3


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    Science Advisor

    Yes, we know what integration by parts is! How about showing exactly what you did so we can point out any mistakes?

    Last edited by a moderator: May 3, 2017
  5. Sep 30, 2008 #4
    I have attached my work with this post. I felt that I was heading in the right direction, but obviously I must have done something wrong near the beginning I would assume, since by the end I found that my integral diverged.

    Attached Files:


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