<s>[INST] 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?
 [/INST] Method by Fourier Series
Consider the function $f(x) = x(1 - x)$, $0 \le x \le 1$. It has Fourier sine series expansion
$$f(x) = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{1}{(2n-1)^3}\sin{(2n-1)\pi x}.$$
Setting $x = \frac{1}{2}$ results in 
$$\frac{1}{4} = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3},$$
or 
$$\frac{\pi^3}{32} = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}.$$
By reindexing the sum we can write
$$\frac{\pi^3}{32} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$
Method by Contour Integration
Let $g(z) = \frac{1}{(2z - 1)^3}$. Then $g$ has only one pole of order $3$ at $z = \frac{1}{2}$. Let $N$ be a positive integer, and consider the contour integral
$$\frac{1}{2\pi i}\int_{\Gamma_N} \pi\csc \pi z\, g(z)\, dz,$$
where $\Gamma_N$ is a positively oriented square with vertices at $\left(N + \frac{1}{2}\right)(\pm 1 \pm i)$. The residue theorem gives
\begin{align}\frac{1}{2\pi i}\int_{\Gamma_N} \pi \csc \pi z\, g(z)\, dz &= \sum_{n = -N}^N \operatorname{Res}\limits_{z = n} \pi \csc \pi z\, g(z) + \operatorname{Res}\limits_{z = \frac{1}{2}} \pi \csc \pi z\, g(z)\\
&= \sum_{n = -N}^N (-1)^n g(n) + \frac{\pi^3}{16}.
\end{align}
For $|z| \ge 1$, $|g(z)| \le |z|^{-3}$. Thus, $$\frac{1}{2\pi i}\int_{\Gamma_N} \pi \csc \pi z\, g(z)\, dz \to 0 \quad \text{as} \quad N \to \infty.$$ 
Hence
$$0 = \sum_{n = -\infty}^\infty (-1)^n g(n) + \frac{\pi^3}{16}$$
that is, 
$$\frac{\pi^3}{16} = \sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}.$$
Now
\begin{align}\sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3} &= \sum_{n = -\infty}^0 \frac{(-1)^{n-1}}{(2n-1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\
& = \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n+1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\
& = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.
\end{align}
Thus
$$\frac{\pi^3}{16} = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$
Finally, we have 
$$\frac{\pi^3}{32} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$
</s>