Evaluate $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} $ How do you evaluate: 
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$$
Not too sure where to start. I thought of trying to convert the denominator into some integral, but that didn't seem to work. According to Wolfram Alpha, the answer is $\frac{\pi^2}{32} $ but I'm not sure how. There's some use of a generalized zeta function on the Wolfram Alpha page, but I'm not too sure how to work with that. If someone would give me the answer using zeta functions, could you show step by step? I would really appreciate it. If there's an easier way, that would be appreciated! 
Oh Also, please no Complex Analysis! Thanks in advanced!
 A: Since $\frac{1}{(2n+1)^3} = \frac{1}{2}\int_{0}^{1} x^{2n}\log^2(x)\,dx$, the whole series can be written as
$$ \frac{1}{2}\int_{0}^{1}\frac{\log^2(x)}{1+x^2}\,dx \stackrel{x\mapsto 1/x}{=}\frac{1}{4}\int_{0}^{+\infty}\frac{\log^2(x)}{1+x^2}\,dx$$
and evaluated in many ways. If you want to avoid Complex Analysis or differentiation under the integral sign at all costs, you may simply enforce the substitution $x=\tan \theta$, then apply Parseval's theorem to the Fourier series of $\log(\tan\theta)=\log(\sin\theta)-\log(\cos\theta)$. The outcome is
$$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3} = \frac{\pi}{4}\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^3}{32}. $$
Anyway, this thread shows that the series $\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{2k+1}}$ are simply related to Euler numbers.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{3}} & =
-\ic\sum_{n = 0}^{\infty}{\ic^{2n + 1} \over \pars{2n + 1}^{3}} =
-\ic\sum_{n = 1}^{\infty}{\ic^{n} \over n^{3}}\,{1 - \pars{-1}^{n} \over 2} =
\Im\sum_{n = 1}^{\infty}{\ic^{n} \over n^{3}}
\\[5mm] & =
\Im\,\mrm{Li}_{3}\pars{\expo{\ic\pi/2}} =
\Im\,\mrm{Li}_{3}\pars{\exp\pars{2\pi\ic\,{1 \over 4}}} =
{1 \over 2\ic}\
\overbrace{\bracks{-\,{\pars{2\pi\ic}^{3} \over 3!}
\,\mrm{B}_{3}\pars{1 \over 4}}}
^{\ds{Jonqui\grave{e}re\ Inversion\ Formula}}
\\[5mm] & =
{2 \over 3}\,\pi^{3}\,\mrm{B}_{3}\pars{1 \over 4} =
{2 \over 3}\,\pi^{3}\bracks{\pars{1 \over 4}^{3} -
{3 \over 2}\,\pars{1 \over 4}^{2} + {1 \over 2}\,\pars{1 \over 4}}
\\[5mm] & = \bbx{\pi^{3} \over 32} \approx 0.9689
\end{align}

$\ds{\mrm{Li}_{s}}$ is the
  Polylogarithm Function. $\ds{\,\mrm{B}_{s}}$ is a Bernoulli Polynomial. See
  Jonqui$\mrm{\grave{e}}$re Inversion Formula.

$$
\mbox{Note that}\quad
\mrm{B}_{3}\pars{x} = x^{3} - {3 \over 2}\,x^{2} + {1 \over 2}\,x
\phantom{\mbox{Note that}\quad}
$$
