I would like to see the ways that you prove that $$\sum_{n\ge0}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}.$$All methods are welcome.
My Proof:
We start by breaking up the sum as $$S=\sum_{n\ge0}\frac{(-1)^n}{(2n+1)^3}=\sum_{n\ge0}\frac1{(4n+1)^3}-\sum_{n\ge0}\frac1{(4n+3)^3}$$ and seeing that this can be expressed in terms of the polygamma function: $$S=\frac1{2^7}\left[\psi_2\left(\frac34\right)-\psi_2\left(\frac14\right)\right].$$ Then using $$\psi_s(z)=\int_0^1\frac{t^{z-1}}{t-1}\ln^s(t)dt$$ we have $$S=\frac{1}{2^7}\int_0^1\frac{\ln^2(x)}{x-1}(x^{-1/4}-x^{-3/4})dx.$$ Then taking advantage of $x\mapsto 1/x$, $$S=\frac1{2^7}\int_1^\infty \frac{\ln^2(x)}{1-x}\cdot\frac{1-x^{1/2}}{x^{3/4}}dx.$$ Then $x\mapsto x^4$ cleans this up quite nicely: $$S=\frac12\int_1^\infty \frac{1-x^2}{1-x^4}\ln^2(x)dx=\frac12\int_1^\infty \frac{\ln^2(x)}{x^2+1}dx.$$ Then we use a little symmetry: $$S=\frac14\int_0^\infty \frac{\ln^2(x)}{x^2+1}dx.$$ Finally, we use $x\mapsto\tan x$: $$S=\frac14\int_0^{\pi/2}\ln^2(\tan x)dx.$$ We immediately recognize this as $$\int_0^{\pi/2}\ln^2(\tan x)dx=\left(\frac{\partial}{\partial a}\right)^2\frac{\Gamma(\frac{1+a}2)\Gamma(\frac{1-a}{2})}2\bigg|_{a=0}$$ but the Gamma reflection formula gives this as $$\frac\pi2\left(\frac{\partial}{\partial a}\right)^2\frac{1}{\cos\frac{\pi a}{2}}\bigg|_{a=0}=\frac{\pi^3}{8},$$ which provides our result.
