Question on a Basel-like sum Hello all,
I have happened upon the following sum:
$ 1^2 + \Big(1 \times  \frac{1}{3} + \frac{1}{3} \times 1 \Big)^2 + \Big(1 \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{5} \times 1 \Big)^2  +  \Big(1  \times \frac{1}{7} + \frac{1}{3}  \times \frac{1}{5} + \frac{1}{5}  \times  \frac{1}{3} + \frac{1}{7}  \times 1 \Big)^2 + \dots  = \frac{\pi^4}{32} $
However, the proof I know is fairly nonilluminative: it more or less just falls from the sky. I'm wondering whether anyone can see another way to prove it, in particular whether it is implied by Euler's Basel sum for odd integers:
$ 1+\frac{1}{3^2} + \frac{1}{5^2} + \ldots = \frac{\pi^2}{8}$
The new sum looks like some sort of convolution of Euler's sum, and the value of the first is double the square of the value of the second. If anyone can shed any light, I'll be much obliged.
P.S. For the proof known to me, see
http://www.tandfonline.com/doi/abs/10.1080/10652469.2012.689301
http://arxiv.org/abs/1205.2458

Interesting problem.  Here is my version.
$$\begin{align}
f(u) &= u+ \frac{u^3}{3}+\frac{u^5}{5}+\dots = \frac{1}{2}\log\frac{1+u}{1-u}
\cr
f(u)^2 &= u^2 + \left(1\cdot\frac{1}{3}+\frac{1}{3}\cdot 1\right)u^4 +
\left(1\cdot \frac{1}{5}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{5}\cdot 1\right)u^6+\dots
\end{align}$$
out of time now, more later...
added
OK, Noam has now done much of my intended solution.  Continuing (avoiding $L$ which I don't know):
$$
\frac{1}{2\pi}\int_{-\pi}^{\pi} f(e^{ix})^2 f(e^{-ix})^2 dx = 
1^2 + \left(1\cdot\frac{1}{3}+\frac{1}{3}\cdot 1\right)^2 +
\left(1\cdot \frac{1}{5}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{5}\cdot 1\right)^2+\dots
=A
$$
So by trigonometry and symmetry
$$
A=\frac{1}{128\pi}\int_{0}^{\pi/2} \left(\log\left(\cot^2\frac{x}{2}\right)^2+
\pi^2\right)^2dx
$$
change variables $z=\log(\cot(x/2))$ to get
$$
A = \frac{1}{128\pi}\int_0^\infty(4z^2+\pi^2)^2 \mathrm{sech} z dz
$$
Then consult Gradsteyn & Ryzhik 3.532 for identities
$$
\int_0^\infty \mathrm{sech}z dz = \frac{\pi}{2},
\int_0^\infty z^2\mathrm{sech}z dz = \frac{\pi^3}{8},
\int_0^\infty z^4\mathrm{sech}z dz = \frac{5\pi^5}{32},
$$
which can be plugged in to get
$$
A = \frac{\pi^4}{32} .
$$
