The more I see integrals resolved with polygamma, polylogarithm and their identities, the more I realize how much I have forgotten over the years. Good refresher for me!
An elegant method to calculate this integral is to use the residue theorem . You can take a semicircle of radius R in the complex plane , with the diameter on the real axis . If let R-> inf only the part on the real line remains.All you have to do is to calculate the residues corresponding to the zeros exp[ i π /8] in the upper halfplane . 2 π i *(residues ) = value of the integral =π /2*√(2+ √2 )
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
WOW!!!
We always learn something different when solving integrals in this channel. thanks!
You are welcome
Awesome 💯
Thanks
The more I see integrals resolved with polygamma, polylogarithm and their identities, the more I realize how much I have forgotten over the years. Good refresher for me!
I am glad you likes it.
Polygamma + Lerch = 🔥🔥
An elegant method to calculate this integral is to use the residue theorem . You can take a semicircle of radius R in the
complex plane , with the diameter on the real axis . If let R-> inf only the part on the real line remains.All you have to do is
to calculate the residues corresponding to the zeros exp[ i π /8] in the upper halfplane . 2 π i *(residues ) = value
of the integral =π /2*√(2+ √2 )
I like it.