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?

Here is another way, and it combines integration and probability.
First off, consider the triple integral:
$$I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^2y^2z^2}dzdydx.$$
Convert this integrand into a geometric series:
$$\frac{1}{1+x^2y^2z^2}=\sum_{n=0}^{\infty} (-1)^n(xyz)^{2n}.$$
Replace the integrand with this series and integrate term by term to get that: $$I=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}.$$ 
Now we proceed to evaluate $$I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^2y^2z^2}dzdydx.$$ Make the change of variables $$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(w)},z=\frac{\sin(w)}{\cos(u)}$$ which has a nice Jacobian $\frac{\partial(x,y,z)}{\partial(u,v,w)}=1+x^2y^2z^2$ that cancels with the integrand. The region of integration is the open polytope $U$ described by inequalities $$0<u+v<\frac{\pi}{2},0<v+w<\frac{\pi}{2},0<u+w<\frac{\pi}{2}, u,v,w>0.$$ We need to compute the volume of $U$ to get the value of $I$.
For the purpose of this proof, we consider the scaled polytope $V$ defined by inequalities:  $$0<u+v<1,0<v+w<1,0<u+w<1, u,v,w>0.$$ Notice $$\text{Vol}(U)=\left(\frac{\pi}{2}\right)^3 \text{Vol}(V)$$ where $\text{Vol}$ means volume. 
We compute $\text{Vol}(V)$ through probability.
Suppose $n=(n_1,n_2,n_3) \in V.$ We first intend to find the probability: $$\text{Pr}\left(n\in V \cap \text{each } n_i <\frac{1}{2}\right)$$ It turns out that:
$$\text{Pr}\left(n\in V \cap \text{each } n_i <\frac{1}{2}\right)=\left(\frac{1}{2}\right)^3=\frac{1}{8}.$$ This is the case because $n$ would lie in the open hypercube $\left(0,\frac{1}{2}\right)^3,$ and one can verify that $\left(0,\frac{1}{2}\right)^3 \subset V.$
Next, we intend to find the probability: $$\text{Pr}\left(n\in V \cap \text{exactly one } n_i \geq \frac{1}{2}\right)$$ It turns out that:
$$\text{Pr}\left(n\in V \cap \text{exactly one } n_i \geq \frac{1}{2}\right)=3\int_{\frac{1}{2}}^{1}\int_{0}^{1-x}\int_{0}^{1-x}=\frac{1}{8}.$$
Here, to get this answer, I computed the probability $\text{Pr}\left(n\in V \cap  n_1 \geq \frac{1}{2}\right)$ and multiplied this answer by $3$ to account for all the possible $3$ cases of this event happening. 
Notice that we cannot have more than one $n_i \geq \frac{1}{2}$ at the same time, as this will violate the constraints of $V$. This means we add up the computed probabilities to get that : $$\text{Vol}(V)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}.$$ This means that 
$$I=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\text{Vol}(U)=\left(\frac{\pi}{2}\right)^3 \text{Vol}(V)=\left(\frac{\pi^3}{8}\right)\frac{1}{4}=\frac{\pi^3}{32}.$$ 
