A wonderful generalised integration result using Feynman's trick

[@maths_505]

If you like the videos and would like to support the channel: www.patreon.com/Maths505 You can follow me on Instagram for write ups that come in handy for my videos: instagram.com/maths.505?igshid=MzRlODBiNWFlZA==

[@MrWael1970]

Thank you for this stunning solution.

[@kkanden]

can't wait to flex on other people with the party trick

[@fartoxedm5638]

Such a wonderful result! For some reason I was sure that we want to evaluate the resulting integral at a = 1(bruh) so I dropped this idea and started regular series expansion analysis. I did the substitution x = e^u, than divided it into parts (-oo, o), (0, oo) and solved them using series expansion of 1/(1 + x) (very uncommon method by the way!) And the result was.... quite nice! It gave me the series in terms of (-1)^(n + 1) / (1 + n * s)^2 where n goes from -oo to oo So by accident I've discovered that sum((-1)^(n + 1) / (1 + n * s)^2, -oo, oo) = -pi^2 / s^2 * csc(pi/s) * cot(pi / s)... And I'm pretty satisfied with that result, you saved my evening. Thanks! Also as far as my strong (desmos) analysis can tell they are equal at every point(not only at (1, oo))!!!!

[@fartoxedm5638]

I want to believe it was not so obvious....

[@maths_505]

Nice work bro

[@fartoxedm5638]

For those who interested, I was able to prove that pi^2 / s^2 * csc^2(pi / s) = sum(1/(1 + n * s)^2, -oo, oo) (it implies from taking the derivative of cot(pi/s) series(Euler one))). So we have another beautiful result that sum(1/(1 + n * s)^2, -oo, oo) * cos(pi/s) = sum((-1)^n / (1 + n * s)^2, -oo, oo)!

[@maths_505]

Beautiful indeed

[@slavinojunepri7648]

Cool result Indeed

[@danielc.martin]

I have a question: for f convergent in the next situation, lim n->infinity integral(from -inf to inf) of (f) /(1+x^n)? Thanks

[@nicogehren6566]

very nice

[@manstuckinabox3679]

It came to the point I can do these things mentally now. thank you Mathematics five-hundred fifty five. btw: both these integrals are what the cool kids call "A Gamelin Classic". found on the chapter of Contour integration with multi-valued functions.

[@appybane8481]

505=five-hundred and five

[@manstuckinabox3679]

@@appybane8481 Yeah the classes were full, so I had to take Maths 555 instead.

[@maths_505]

@@manstuckinabox3679 😂😂😂

[@jiahao2709]

are you doing this in your mobile or?

[@maths_505]

Yup

[@maths_505]

Still on the phone...

[@thomasblackwell9507]

This looks familiar to one of your previous posts or maybe I am just going senile. You never can tell with all the math!

[@maths_505]

It is exactly that post but I messed up the result at the end so I wanted to redo it

[@thomasblackwell9507]

@@maths_505 Thanks, I thought that I was losing my mind “again” when I saw this post. I appreciate the quick response.

[@maths_505]

@@thomasblackwell9507 my friend we're interested in tough calculus problems for applied math.... I'm pretty sure we both lost sanity a long while back😂

[@thomasblackwell9507]

@@maths_505 Then it is nice to know that I am in good company, I thank you.

[@aravindakannank.s.]

I was about to comment on no video again

[@yoav613]

Noice and easy

[@giuseppemalaguti435]

I=I'(0)...I(a)=1/sB(a+1/s,1-a+1/s)....but i dont know the derivate....with geometric series I=(2s-s^2)S[(-1)^k(2k+1)/((ks+1)((k+1)s-1))^2] ,s>1..poi non riesco più a semplificare

[@jeffreyrodrigoecheverria2613]

I got the answer to be: (π^2 (sec^2(π/(2s)) - csc^2(π/(2s))))/(4*s^2)

[@robertsandy3794]

Have you considered renaming your channel to something like: Gamma Boi Gammera Gamma Beta .......

[@maths_505]

😂😂😂 That would be injustice to all the series expansions I use😂