# Recent content by seanieg89

5. ### Higher Level Integration Marathon & Questions

I=\int_0^{2\pi} (\frac{e^{it}+e^{-it}}{2})^{2n}\, dt \\ \\ = 2^{-2n}\binom{2n}{n}\int_0^{2\pi}\, dt\\ \\=\frac{\pi}{2^{2n-1}}\binom{2n}{n}.\\ \\ $(all nonconstant trig monomials integrate to zero.)$
6. ### Higher Level Integration Marathon & Questions

Fair enough, seems a rather subjective matter...to me polylog and polygamma manipulations are okay, but I also have no qualms about using contour integration etc.
7. ### Higher Level Integration Marathon & Questions

Eh I wouldn't say that necessarily, I did the integral in a not particularly imaginative way...there could be a clever shortcut. To me it does seem like computing this integral is roughly the same mathematical "depth" as computing some particular values of the di/trilogarithm though. Anyway...
8. ### Higher Level Integration Marathon & Questions

^ That was good practice! I need to go out for dinner v soon, so will supply gory details later. In short, the answer can be expressed solely in terms of Apery's constant zeta(3). 1. IBP to essentially reduce the integral to log^2(1+x)/x. (The bulk of the difficulty of the originally posted...

Gg man, pretty well played. Shame about that pawn blunder.

Someone has joined, but just isn't moving lol.

Coming? Otherwise someone else join pls :).