Which means that Hω at t equal to 0 right, this is identically equal to half because the second integral that vanishes right. So, Hωo at t equals to 0 is half right. And of course, what I have is that if I my function f of omega 0 evaluated at time point 0 so, this is my point of discontinuity if you recall. So, this is my point of discontinuity.
Hω का मान t = 0 पर 1⁄2 है क्योंकि दूसरा इंटीग्रल लुप्त हो जाता है |यदि मेरा फंक्शन fω0 का मान बिंदु t=0 पर ज्ञात किया जाता है , तो याद कीजिए यह मेरा पॉइंट ऑफ डिस्कंटीन्यूटी है |