TITLE: Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$
QUESTION [7 upvotes]: I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$. 
I had given the following solution:
$$\begin{aligned} 
1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\  
 &\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\  
 &=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\  
 &= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\ 
 &=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4}  \right ] \\ 
 &=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32} 
\end{aligned}$$
where I used polygamma identities and made use of the absolute convergence of the series at $(*)$ in order to re-arrange the terms. 
Any other approach using Fourier Series, or contour integration around a square, if that is possible?

REPLY [4 votes]: This is $\beta(3)$, where $\beta(x)$ is the Dirichlet Beta Function. The derivation below is taken from this answer.

The Dirichlet Beta Function
$$
\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\tag1
$$
is known as the Dirichlet beta function. As shown below,
$$
\beta(3)=\dfrac{\pi^3}{32}\tag2
$$

Recurrence for $\boldsymbol{\beta(2n+1)}$
We can develop a recurrence for $\beta(2k+1)$. First, the generating function for $\beta(2k+1)$ is
$$
\begin{align}
f(x) &= \sum_{k=0}^\infty \beta(2k+1) x^{2k+1}\\
     &= \sum_{n=0}^\infty \sum_{k=0}^\infty (-1)^n\left(\frac{x}{2n+1}\right)^{2k+1}\\
     &= \sum_{n=0}^\infty (-1)^n\frac{\frac{x}{2n+1}}{1-\left(\frac{x}{2n+1}\right)^2}\\
     &= \frac{x}{2} \sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+x}+\frac{1}{2n+1-x}\right)\\
     &= \frac{x}{2} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{2n+1+x}\\
     &= \frac{x}{4} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{n+\tfrac{1+x}{2}}\\
     &= \frac{x}{4} \pi \csc\left(\pi\tfrac{1+x}{2}\right)\\[3pt]
     &= \frac{\pi}{4} x \sec\left(\frac{\pi}{2}x\right)\tag{3}
\end{align}
$$
where we use $(7)$ from this answer to get
$$
\begin{align}
\sum_{n=-\infty}^{+\infty}\frac{(-1)^n}{n+z}
&=\sum_{n=-\infty}^{+\infty}\frac2{2n+z}-\sum_{n=-\infty}^{+\infty}\frac1{n+z}\\[3pt]
&=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt]
&=\pi\csc(\pi z)\tag{4}
\end{align}
$$
We can use equation $(3)$ to develop a recurrence relation:
$$
\begin{align}
\frac{\pi}{4} x
&= \cos\left(\frac{\pi}{2} x\right) f(x)\\
&= \sum_{n=0}^\infty\sum_{k=0}^n (-1)^k \frac{(\frac{\pi}{2} x)^{2k}}{(2k)!}\;\beta(2n-2k+1)x^{2n-2k+1}\\
&= \sum_{n=0}^\infty\sum_{k=0}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)x^{2n+1}\tag{5}
\end{align}
$$
For $n=0$, we can use the arctan series to get
$$
\beta(1) = \frac{\pi}{4}\tag{6}
$$
and for $n\gt0$, $(5)$ gives
$$
\beta(2n+1) = -\sum_{k=1}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)\tag{7}
$$
Recursion $(7)$ yields
$$
\begin{align}
\beta(1)&=\frac{\pi}{4}\\
\beta(3)&=\frac{\pi^3}{32}\\
\beta(5)&=\frac{5\pi^5}{1536}\\
\beta(7)&=\frac{61\pi^7}{184320}\\
\beta(9)&=\frac{277\pi^9}{8257536}\\
\beta(11)&=\frac{50521\pi^{11}}{14863564800}\\
\beta(13)&=\frac{540553\pi^{13}}{1569592442880}\\
\beta(15)&=\frac{199360981\pi^{15}}{5713316492083200}\\
\beta(17)&=\frac{3878302429\pi^{17}}{1096956766479974400}\\
\beta(19)&=\frac{2404879675441\pi^{19}}{6713375410857443328000}
\end{align}
$$