An amazing calculus result derived using contour integration

Here is an awesome integral that I've previously evaluated using real analytic methods and this time we're using contour integration which itself provides a cool way of deriving such an elegant result.

[hanspetermarro]

I like your clips and this one is very nice again.
There is, however, a small mistake after 12:42 where you state the supposed reason for the integral along Gamma going to zero. While it is correct that the expression exp(-R sin(phi)) approaches zero as R become large, this and the fact that sin(phi)>0 for 0<phi<pi mean that the numerator of the integrand, which is the _exponential_ of exp(-R sin(phi))cos(...), approaches 1 when 0<phi<pi. (The circumstance that this does not happen at phi=0 and phi=pi does not affect the value of the integral). The acutal reason for the "collapse" of the integral is the factor R in the denominator.

[jieyuenlee]

13:47
Im(e^e^ix)=Im(e^cosx e^isinx)
=e^cosx×Im(e^isinx)
=e^cosx×sin(sinx)
This is an odd function
Hence Imaginary part of int(e^e^ix)/(x²+1)dx from(-inf, inf)=0

[GreenMeansGOF]

So the sin(sin(x)) version is the imaginary part which just equals 0?

[zahari]

This is Problem 2116 from the Mathematics Magazine.The solution is in vol 95(2), 2022, p. 158.

[manstuckinabox]

HEY IT'S MY FAV FUNCTION DEVIDED BY MY FAV DENOMINATOR!!

[cadmio]

Everything can be contour integrated if you are brave enough

[maalikserebryakov]

Contour integration is an absolutely broken method. Thankfully Mathematics can’t be nerfed.

[wizz_larza]

i'll be honest, i hate contour integration 😢