I was trying to evaluate $$I=\int_0^{\infty} \frac{\cos x}{e^x+e^{-x}}\text{d}x,$$ and am trying to find my error. Here's my work, with tons of algebra left out.
$$I=\frac{1}{2}\int_0^{\infty}\frac{e^{ix}+e^{-ix}}{e^x+e^{-x}}\text{d}x$$
$$=\frac{1}{2}\int_0^{\infty}\left(e^{(1+i)x}+e^{(1-i)x}\right)\sum_{n=0}^{\infty}(-1)^nx^{2n}\text{d}x$$
$$=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\int_0^{\infty}x^{2n} e^{(1+i)x}\text{d}x+\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\int_0^{\infty}x^{2n} e^{(1-i)x}\text{d}x$$
On those two integrals I do $u=-(1+i)x$ and $u=-(1-i)x$, respectively:
$$=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\frac{-1}{(1+i)^{2n+1}}\int_0^{\infty}u^{2n} e^{-u}\text{d}u+\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\frac{-1}{(1-i)^{2n+1}\int_0^{\infty}u^{2n} e^{-u}}\text{d}u$$
$$=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\frac{-1}{(1+i)^{2n+1}}\Gamma(2n)+\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\frac{-1}{(1-i)^{2n+1}}\Gamma(2n)$$
At this point I know I've screwed up, because these sums don't diverge. But I'll show my last steps:
$$=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^{n+1}(2n-1)!\frac{1}{(1+i)^{2n+1}+\frac{1}{(1-i)^{2n+1}}}.$$
Omitting some complex numbers work:
$$=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^{n+1}(2n-1)!\frac{-\sqrt{2}\cos(\frac{\pi}{4}(2n+1))}{2^n}$$
And this certainly diverges because $\lim_{n\to\infty}\frac{(2n-1)!}{2^n}=\infty$...
Someone help!
