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?
 A: 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 
$$\frac{\pi^3}{32} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$
A: Differentiating twice the logarithm of the Weierstrass representation of sine gives
$$ \sum\limits_{n=-\infty}^{\infty} {1\over (z+n)^2}=\frac{\pi^{2}}{\sin^{2}(\pi z)} $$
(as i've been answered in here.)
Now differentiate once more and consider $z=\frac{1}{4}$.
A: I do not know how much this could help you; so forgive me if I am out off topic.
Rewriting a little the expression $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}x^n=\frac{1}{8} \,\Phi \left(-x,3,\frac{1}{2}\right)$$ where appears the Lerch transcendent function. Now, using $x=1$, we can get the result.
A: 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 $\int_0^{\pi/2}\tan^{2k-1}tdt=\frac{\pi}{2}\csc(k\pi)$,$$S=\frac{\pi}{32}\left.\left(\partial_k^2\csc(k\pi)\right)\right|_{k=\frac12}=\frac{\pi^3}{32}\left(\csc\frac{\pi}{2}\right)\left(\cot^2\frac{\pi}{2}+\csc^2\frac{\pi}{2}\right)=\frac{\pi^3}{32}.$$
A: 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 
$$I=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\text{Vol}(U)=\left(\frac{\pi}{2}\right)^3 \text{Vol}(V)=\left(\frac{\pi^3}{8}\right)\frac{1}{4}=\frac{\pi^3}{32}.$$ 
