# 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...

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 $$A = \frac{\pi^4}{32} .$$

-
[sorry, couldn't tell from just the first two lines that this was basically the same as what I was doing. BTW, seems like there might be a factor of $4$ missing in front of $\log(\cot^2 \frac{x}{2})^2$.] – Noam D. Elkies Jul 25 '12 at 7:42
Maple gives me this numerical value: (1/128)*(int((ln(cot((1/2)*x)^2)^2+Pi^2)^2, x = 0 .. (1/2)*Pi))/Pi = 3.044034094 – Gerald Edgar Jul 25 '12 at 15:40
@Noam: yes, this must be closely connected to yours. G & R says that the integrals $\int_0^\infty z^{2m} \mathrm{sech} z dz$ are Euler numbers time powers of $\pi$. – Gerald Edgar Jul 25 '12 at 15:43
The numerical value agrees with $\pi^4/32$. The integrals are the same as those I obtained; I just wrote sech in terms of exponentials. You can get a generating function for these integrals from the Fourier transform of the hyperbolic secant. (And I still think there's a missing factor of 4 in front of $\log(\cot^2(x/2))$ in that integral.) – Noam D. Elkies Jul 25 '12 at 16:45
Thank you to both Professors Elkies and Edgar for these interesting answers. Not surprisingly, my method uses the coefficients for the same map, but notes that it is a conformal map from the unit disk onto the interior of a parabola, and then solves a particular PDE(very easy on the parabola) which gives the solution to the integrals in question. I am happy to see these alternate methods, though. – Greg Markowsky Aug 8 '12 at 2:04

[Edited to fill in a couple of steps, correct some typos, standardize the $L$-function notation, and add comments about the relation with the "Basel sum"]

Here's a proof via transformation of the sum to $$\phantom{(*)CCCCCCCCCCCC} \left(\frac\pi4\right)^4 + \frac\pi4 L(3,\chi_4) + \frac6\pi L(5,\chi_4), \phantom{CCCCCCCCCCCC}(*)$$ where $L(\cdot,\chi_4)$ is the Dirichlet $L$-function defined for $s>0$ by $$L(s,\chi_4) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} = 1 - \frac1{3^s} + \frac1{5^s} - \frac1{7^s} + - \cdots .$$ Since it is known that $L(3,\chi_4) = \pi^3/32$ and $L(5,\chi_4)=5\pi^5/1536$, this formula comes to $$\left(\frac1{256} + \frac1{128} + \frac5{256}\right) \pi^4 = \frac{\pi^4}{32}$$ as desired.

It was observed already (implicitly by the proposer, and now as I see also explicitly in G.Edgar's incipient answer [later completed]) that the numbers $c_2=1$, $c_4=1\times\frac13 + \frac13\times 1$, $c_6=1\times\frac15 + \frac13\times\frac13 + \frac15\times1$, etc. are the coefficients of $x^2,x^4,x^6,\ldots$ in the power-series expansion of $f(x)^2$ where $$f(x) = \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1} = \frac12 \log\frac{1+x}{1-x}.$$ Hence the desired sum $c_2^2 + c_4^2 + c_6^2 + \cdots$ is $(2\pi)^{-1} \int_{-\pi}^\pi |f(e^{it})|^4 dt$ by Parseval's formula. Now for $0<|t|<\pi$ we have $$f(e^{it}) = \frac12 \log\frac{1+e^{it}}{1-e^{it}} = \frac12 \log \left| \cot \frac{t}{2} \right| + \frac{\pi i}{4} \mathop{\rm sgn}t,$$ whence $$\left|f(e^{it})\right|^2 = \frac14 \log^2 \left| \cot \frac{t}{2} \right| + \frac{\pi^2}{16}$$ which is symmetrical under $t \leftrightarrow -t$. Since $\cot(t/2) > 0$ for $0<t<\pi$, we're left with the task of evaluating $$\frac1\pi \int_0^{\pi} \left( \frac14 \log^2 \cot \frac{t}{2} + \frac{\pi^2}{16} \right)^2 dt,$$ which is to say $$\frac1{16\pi} \int_0^{\pi} \log^4 \cot \frac{t}{2} dt + \frac\pi{32} \int_0^{\pi} \log^2 \cot \frac{t}{2} dt + \frac{\pi^4}{4^4}.$$ So change variables to $u = \log \cot(t/2)$, which ranges from $+\infty$ to $-\infty$ as $t$ goes from $0$ to $\pi$. We calculate $dt = -2 e^u du/(1+e^{2u})$, so for each $m=0,1,2,\ldots$ we have $$\int_0^{\pi} \log^{2m} \cot \frac{t}{2} dt = 2 \int_{-\infty}^\infty u^{2m} \frac{e^u}{1+e^{2u}} du = 4 \int_0^\infty u^{2m} \frac{e^u}{1+e^{2u}} du,$$ using in the last step the symmetry of the integrand under $u \leftrightarrow -u$. Expanding $e^u/(1+e^{2u})$ as $e^{-u} - e^{-3u} + e^{-5u} - e^{-7u} + - \cdots$, multiplying each term by $u^{2m}$, and integrating yields $$\int_0^{\pi} \log^{2m} \cot \frac{t}{2} dt = 4 (2m)! \left( 1 - \frac1{3^{2m+1}} + \frac1{5^{2m+1}} - \frac1{7^{2m+1}} + - \cdots \right) = 4 (2m)! L(2m+1,\chi_4).$$ Hence our integral equals (*), QED.

Remarks: i) The term $(\pi/32) \int_0^{\pi} \log^2 \cot (t/2) dt$ could also have been evaluated by applying Parseval to the Fourier series of $\log \cot (t/2)$, but I do not see how to get at $\int_0^{\pi} \log^4 \cot (t/2) dt$ in this way.

ii) The proposer notes that the sum $c_2^2 + c_4^2 + c_6^2 + \cdots = \pi^4/32$ "looks like some sort of convolution of Euler's sum" $1 + (1/3^2) + (1/5^2) + (1/7^2) + \cdots = \pi^2/8$, but the evaluation of $\sum_{m=1}^\infty c_{2m}^2$ does not seem to follow directly from this formula, even though all the steps in the above derivation were available to Euler (indeed the values of $L(2m+1,\chi_4)$ are expressed in terms of Euler numbers!). For example, the same "sort of convolution" applied to $1 + (1/3^4) + (1/5^4) + (1/7^4) + \cdots = \pi^4/96$ yields a sum that does not (as far as I can see) yield to the same technique, and might not even be a rational multiple of $\pi^8$.

-