TITLE: Infinite Series $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$
QUESTION [17 upvotes]: I'm looking for a way to prove
$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$
I know that
$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$
so maybe I could simplify the above more?

REPLY [2 votes]: \begin{gather*}
\beta(2m+1)=\frac{1}{(2m)!}\int_0^1\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x=\frac{1}{(2m)!}\left(\int_0^\infty-\int_1^\infty\right)\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x\\
=\frac{1}{(2m)!}\int_0^\infty\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x-\frac{1}{(2m)!}\underbrace{\int_1^\infty\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x}_{x\to 1/x}\\
=\frac{1}{(2m)!}\int_0^\infty\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x-\frac{1}{(2m)!}\int_0^1\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x\\
\left\{\text{add  $\beta(2m+1):=\frac{1}{(2m)!}\int_0^1\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x$ to both sides then divide by $2$}\right\}\\
=\frac{1}{2(2m)!}\int_0^\infty\frac{\ln^{2m}(x)}{1+x^2}\mathrm{d}x\\
=\frac{\pi}{4^{m+1}(2m)!}\lim_{s\to \frac12}\frac{d^{2m}}{ds^{2m}}\csc(\pi s). 
\end{gather*}
We showed here
$$\lim_{s\to \frac12}\frac{d^{2m}}{ds^{2m}}\csc(s\pi)=|E_{2m}|\pi^{2m}.$$
Using this result, it follows that
$$\beta(2m+1)=\frac{|E_{2m}|\pi^{2m+1}}{4^{m+1}(2m)!}.$$