Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ 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?

We begin with an integral representation:$$S:=\sum_{n\ge0}\frac{(-1)^n}{(2n+1)^3}=\frac12\int_0^\infty\frac{x^2e^{-x}}{1+e^{-2x}}dx.$$The integrand is even, so$$S=\frac14\int_{-\infty}^\infty\frac{x^2e^{-x}}{1+e^{-2x}}dx.$$Or with $e^{-x}=\tan t$,$$S=\frac14\int_0^{\pi/2}\ln^2(\tan t)dt.$$Since $\int_0^{\pi/2}\tan^{2k-1}tdt=\frac{\pi}{2}\csc(k\pi)$,$$S=\frac{\pi}{32}\left.\left(\partial_k^2\csc(k\pi)\right)\right|_{k=\frac12}=\frac{\pi^3}{32}\left(\csc\frac{\pi}{2}\right)\left(\cot^2\frac{\pi}{2}+\csc^2\frac{\pi}{2}\right)=\frac{\pi^3}{32}.$$
