Infinite Series $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$ 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?
 A: The Dirichlet beta function is defined as
$$
\beta(2m+1)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}\tag{1}
$$
Equations $(8)$ and $(9)$ in this answer say that $\beta(1)=\frac\pi4$ and
$$
\beta(2m+1) = -\sum_{k=1}^m \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2m-2k+1)\tag{2}
$$
If we reindex recursion $(7)$ derived below, we get that the even Euler numbers are defined by $\mathrm{E}_0=1$ and
$$
\mathrm{E}_{2m}=-\sum_{k=1}^m\binom{2m}{2k}\mathrm{E}_{2m-2k}\tag{3}
$$
then notice that $(2)$ is the same as $(3)$ if we set
$$
\beta(2m+1)=\frac{(-1)^m\mathrm{E}_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}\tag{4}
$$
QED

Recursion for the even Euler numbers
The Exponential Generating Function for the Euler numbers is $\mathrm{sech}(x)$. This means that the odd Euler numbers are $0$ and
$$
\mathrm{sech}(x)=\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\tag{5}
$$
Therefore,
$$
\begin{align}
1
&=\cosh(x)\,\mathrm{sech}(x)\\[9pt]
&=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{6}
\end{align}
$$
Equation $(6)$ says that $\mathrm{E}_0=1$ and
$$
\mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}\tag{7}
$$
A: My proof works through the following lines: the LHS is:
$$\frac{1}{(2m)!}\int_{0}^{1}\frac{(\log x)^{2m}}{1+x^2}dx = \frac{1}{2\cdot(2m)!}\int_{0}^{+\infty}\frac{(\log x)^{2m}}{1+x^2}dx,$$
so we just need to compute:
$$\left.\frac{d^{2m}}{dk^{2m}}\int_{0}^{+\infty}\frac{x^k}{1+x^2}\right|_{k=0},\tag{1}$$
but:
$$ \int_{0}^{+\infty}\frac{x^{1/r}}{1+x^2}\,dx = r\int_{0}^{+\infty}\frac{y^r}{1+y^{2r}}\,dy = \frac{\pi/2}{\cos(\pi/(2r))}$$
by the residue theorem, so 
$$\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{E_{2m}}{2\cdot(2m)!}\left(\frac{\pi}{2}\right)^{2m+1}\tag{2}$$
where $E_{2m}$ is just the absolute value of an Euler number, that belongs to $\mathbb{N}$.
A: \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
$$\beta(2m+1)=\frac{|E_{2m}|\pi^{2m+1}}{4^{m+1}(2m)!}.$$
A: Because it wasn't explicitly stated, I probably should state that $E_{2m}$ are the Euler numbers.
Let's integrate the function $$ f(z) = \frac{\pi \csc (\pi z)}{(2z+1)^{2m+1}}, \quad m \in \mathbb{N}_{\ge 0},$$
around a square contour with vertices at $\pm (N+\frac{1}{2}) \pm i(N+\frac{1}{2})$ , where $N$ is a positive integer.
Letting $N$ go to infinity throught the positive integers, the integral vanishes (see here), and we end up with
$$ \begin{align} 0 &= \sum_{n=-\infty}^{\infty} \text{Res}[f(z),n] + \text{Res} \left[f(z),- \frac{1}{2} \right] \\ &= 2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}} + \text{Res}\left[f(z),-\frac{1}{2} \right]. \end{align}$$
Notice that
$$\text{Res} \left[ f(z),- \frac{1}{2} \right] = \text{Res} \left[f \left(z - \frac{1}{2}\right),0 \right] = \text{Res} \left[-\frac{\pi \sec (\pi z)}{(2z)^{2m+1}},0 \right] .$$
Using the Maclaurin series of sec(z), that is,  $$ \sec (z) = \sum_{n=0}^{\infty} \frac{(-1)^n E_{2n}}{(2n)!}z^{2n} \, ,  \quad |x| < \frac{\pi}{2},$$
we have
$$-\frac{\pi \sec (\pi z)}{(2z)^{2m+1}} = -\frac{\pi}{2^{2m+1}} \sum_{n=0}^{\infty} \frac{(-1)^{k} E_{2n}}{(2n)!} \pi^{2n} z^{2n-2m-1}  ,$$
from which we can conclude that
$$\text{Res} \left[-\frac{\pi \sec \pi z}{(2z)^{2m+1}},0 \right] = -\frac{\pi (-1)^{m} E_{2m}}{2^{2m+1} (2m)!} \pi^{2m} .$$
Therefore,
$$2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}} = \frac{(-1)^{m} E_{2m}}{2^{2m+1} (2m)!} \pi^{2m+1}\, ,  $$
and the result follows.

EDIT:
Another approach is to use the partial fractions expansion
$$ \sec(z) = \sum_{n=0}^{\infty} \frac{(-1)^{n} (2n+1) \pi}{\left(\frac{2n+1}{2} \right)^{2} \pi^{2}-z^{2}}. $$
For $|z| < \frac{\pi}{2}$, we have
$$ \begin{align}\sec(z) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} (2n+1)\pi}{\left(\frac{2n+1}{2} \right)^{2}\pi^{2}} \frac{1}{1-\left( \frac{2z}{(2n+1)\pi}\right)^{2}} \\ &=\frac{4}{\pi}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \sum_{k=0}^{\infty} \frac{(2z)^{2k}}{\left((2n+1) \pi\right)^{2k}}  \\ &= \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2k+1}} . \end{align} $$
But we know that
$$ \sec(z) = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k} .$$
Therefore,
$$ \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2k+1}} = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k},$$
and the result follows by comparing the $k$th coefficient of both series.
