RECIPROCAL BINOMIAL SUM REPRESENTATION FOR ZETA(4)

Today´s post We will prove the following beautiful result

$\zeta(4)=\frac{36}{17}\sum_{n=1}^\infty\frac{ 1}{n^4\binom{2n}{n}}$

To this end we will go through a lot of results ranging from Logsine integrals and Polylogarithms and Euler Sums.

Let´s start by showing the connection between infinite sums involving the reciprocal of the binomial coefficients and Logsine integrals

First recall the identity (proved here)

$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac12\sum_{n=1}^\infty\frac{ 4^{n}x^{2n-1}}{n\binom{2n}{n}}$ (1)

Letting $x \to \frac{x}{2}$ in (1) we immediately obtain

$\frac{\arcsin\left(\frac{x}{2} \right)}{\sqrt{1-\left( \frac{x}{2}\right)^2}}=\sum_{n=1}^\infty\frac{x^{2n-1}}{n\binom{2n}{n}}$ (2)

Now recall the following integral representation (see here a proof)

$\frac{1}{n^k}=\frac{1}{\Gamma(k)}\int_0^1 \left(-\ln(x) \right)^{k-1}x^{n-1}\,dx$

I want to show that

$\sum_{n=1}^\infty\frac{ 1}{\binom{2n}{n}n^k}=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{\pi/3}x\ln^{k-2}\left(2 \sin\left( \frac{x}{2}\right) \right)\,dx$ (3)

The following proof is based on Borwein (see references below)

Proof:

$\begin{aligned} \sum_{n=1}^\infty\frac{ 1}{\binom{2n}{n}n^k}&=\frac{(-1)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ 1}{\binom{2n}{n}}\int_0^1 \ln^{k-1}(x) x^{n-1}\,dx\\ &=\frac{2(-1)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ 1}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x^2\right) x^{2n-1}\,dx & \left(x \to x^2\right)\\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ 1}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x\right) x^{2n-1}\,dx \\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ 1}{\binom{2n}{n}}\left(\frac{x^{2n}\ln^{k-1}(x)}{2n}\Bigg|_0^1-\frac{(k-1)}{2n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \right)\\ &=-\frac{(-2)^{k-1}}{(k-2)!}\sum_{n=1}^\infty\frac{ 1}{\binom{2n}{n}n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\sum_{n=1}^\infty\frac{ x^{2n-1}}{\binom{2n}{n}n} \right) \,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\frac{ \arcsin\left(\frac{x}{2} \right)}{\sqrt{1-\left( \frac{x}{2}\right)^2}} \right) \,dx & \left( \text{by eq. (2)}\right)\\ &=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{\pi/3} \frac{x\ln^{k-2}\left(2\sin\left(\frac{x}{2}\right)\right) \cos\left(\frac{x}{2} \right)}{\sqrt{1-\sin^2\left( \frac{x}{2}\right)}} \,dx & \left( \frac{x}{2} \to \sin\left(\frac{x}{2} \right)\right)\\ &=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{\pi/3}x\ln^{k-2}\left(2 \sin\left( \frac{x}{2}\right) \right)\,dx \qquad \blacksquare \end{aligned}$

Then according to (3) we can write

$\sum_{n=1}^\infty\frac{ 1}{n^4\binom{2n}{n}}=2\int_0^{\pi/3}x\ln^{2}\left(2 \sin\left( \frac{x}{2}\right) \right)\,dx$ (4)

To evaluate (4), we have to compute the integral on the R.H.S. of (4). As we did before, a good strategy is to use the following expansion for log squared:

$\ln^2\left(2 \sin(x) \right)=\frac{\pi^2}4-\pi x+x^2+2 \sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}\cos(2nx)}{n}$ (5)

proved here.

Note in (5) the last term on the R.H.S.: An infinite series involving the Harmonic numbers and a trigonometric function. Unavoidably we will end up having to compute a nasty infinite sum that stems from this term. So before jumping straight to the computation of the integral in (2), We will evaluate the term that involves the Harmonic numbers and trigonometric functions that will show up in our calculations.

Recall the generating function (see here)

$\sum_{n=1}^\infty \frac{\operatorname{H}_n}{n}x^n=\frac12\ln^2(1-x)+\operatorname{Li}_2(x)$ (6)

Rearraging terms, dividing both sides by x and integrating from 0 to x we obtain

$\begin{aligned} &\sum_{n=1}^\infty \frac{\operatorname{H}_n}{n}\int_0^x t^{n-1}\,dt-\int_0^x\frac{\operatorname{Li}_2(t)}{t}\,dt=\frac12\int_0^x \frac{\ln^2(1-t)}{t}\,dt\\ &\sum_{n=1}^\infty \frac{\operatorname{H}_n}{n^2}x^n-\operatorname{Li}_3(x)=\frac12\int_0^x \frac{\ln^2(1-t)}{t}\,dt\\ &\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}x^n=\frac12\int_0^x \frac{\ln^2(1-t)}{t}\,dt\\ \end{aligned}$

$\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}x^n=\frac12\int_0^x \frac{\ln^2(1-t)}{t}\,dt$ (7)

Dividing both sides of (7) by x and integrating from 0 to x we obtain

$\begin{aligned} \sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^3}x^n&=\frac12\int_0^x \frac{1}{t} \int_0^t \frac{\ln^2(1-u)}{u}\,du\,dt\\ &=\frac12\int_0^x \frac{\ln^2(1-u)}{u} \int_u^x \frac{1}{t}\,dt\,du\\ &=\frac12\int_0^x \frac{\ln^2(1-u)}{u} \left(\ln(x)-\ln(u) \right)\,du\\ &=\ln(x)\frac{1}2\int_0^x \frac{\ln^2(1-u)}{u} \,du-\frac12\int_0^x \frac{\ln^2(1-u)\ln(u)}{u}\,du\\ &=\ln(x)\left(\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}x^n \right)-\frac12\int_0^x \frac{\ln^2(1-u)\ln(u)}{u}\,du \qquad \left(\text{by eq.(2)} \right)\\ \end{aligned}$

Rearranging terms we obtain

$\int_0^x \frac{\ln^2(1-u)\ln(u)}{u}\,du =2\ln(x)\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}x^n -2\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^3}x^n$ (8)

We will be interested in the real part of (8) evaluated at $x=e^{i \pi/3}$ (You can already guess why $\pi/3$ ).

Let´s focus on the R.H.S. of (8) first.

Letting $x=e^{i \pi/3}$ we obtain:

$\begin{aligned} f\left( e^{i \pi/3}\right)&=2\ln(e^{i \pi/3})\left(\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}\left(\cos\left(\frac{n \pi}{3}\right)+i\sin\left(\frac{n \pi}{3} \right) \right)\right)-2\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^3}\left(\cos\left(\frac{n \pi}{3}\right)+i\sin\left(\frac{n \pi}{3} \right) \right)\\ &= \frac{i 2\pi}{3}\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}\cos\left(\frac{n \pi}{3}\right)-\frac{2 \pi}{3}\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}\sin\left(\frac{n \pi}{3} \right) -2\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^3}\cos\left(\frac{n \pi}{3}\right)-2i\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^3}\sin\left(\frac{n \pi}{3} \right) \end{aligned}$

Taking real part we obtain

$\Re\left\{f\left( e^{i \pi/3}\right) \left\}=-\frac{2 \pi}{3}\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}\sin\left(\frac{n \pi}{3} \right)-2\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^3}\cos\left(\frac{n \pi}{3}\right)$ (9)

Let´s now handle with the L.H.S. of (8)

$\begin{aligned} f\left( x}\right)&=\int_0^x \frac{\ln^2(1-u)\ln(u)}{u}\,du \\ &=\frac{\ln^2(1-u)\ln^2(u)}{2}\Big|_0^x+\int_0^x \frac{\ln^2(u)\ln(1-u)}{1-u}\,du & (\text{integration by parts}) \\ &=\frac{\ln^2(1-x)\ln^2(x)}{2}+\int_{1-x}^1 \frac{\ln^2(1-u)\ln(u)}{u}\,du & \left(1-u \to u \right)\\ &=\frac{\ln^2(1-x)\ln^2(x)}{2}+\int_{0}^1 \frac{\ln^2(1-u)\ln(u)}{u}\,du-\int_0^{1-x} \frac{\ln^2(1-u)\ln(u)}{u}\,du & (*)\\ &=\frac{\ln^2(1-x)\ln^2(x)}{2}-\frac{\zeta(4)}{2}-\int_0^{1-x} \frac{\ln^2(1-u)\ln(u)}{u}\,du\\ &=\frac{\ln^2(1-x)\ln^2(x)}{2}-\frac{\zeta(4)}{2}-f(1-x) \end{aligned}$

In (*) we used the result proved here

$\int_{0}^1 \frac{\ln^2(1-u)\ln(u)}{u}\,du=-\frac{\zeta(4)}{2}$

Rearranging terms we get

$f\left( x}\right)+f\left(1- x}\right)=\frac{\ln^2(1-x)\ln^2(x)}{2}-\frac{\zeta(4)}{2}$ (10)

(10) is functional equation for (8)

In general we have

$\mathrm{Re}(f(z)) = \frac{f(z) + \overline{f(z)}}{2}$ (11)

So let´s apply (11) to find the real part on the L.H.S. of (8) using the functional equation (10)

$\begin{aligned} 2\Re\left\{f\left( e^{i \pi/3}\right) \left\}&=f\left( e^{i\pi/3}\right)+f\left(e^{-i\pi/3} \right)\\ &=f\left( e^{i\pi/3}\right)+f\left(1-e^{i\pi/3} \right)\\ &=\frac{\ln^2\left(1-e^{i\pi/3}\right)\ln^2\left(e^{i\pi/3}\right)}{2}-\frac{\zeta(4)}{2} & \left(\text{by eq. (10)} \right)\\ &=\frac{\ln^2\left(e^{-i\pi/3}\right)\ln^2\left(e^{i\pi/3}\right)}{2}-\frac{\zeta(4)}{2}\\ &=\frac12\left(-\frac{i \pi}{3} \right)^2\left(\frac{i \pi}{3} \right)^2-\frac{\zeta(4)}{2}\\ &=\frac12\frac{\pi^4}{81}-\frac{\zeta(4)}{2}\\ &=\frac{\zeta(4)}{18} \qquad \blacksquare \end{aligned}$

Therefore

$2\left(\frac{2 \pi}{3}\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^2}\sin\left(\frac{n \pi}{3} \right)+2\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n^3}\cos\left(\frac{n \pi}{3}\right)\right)=-\frac{\zeta(4)}{18}$ (12)

We are now ready to compute the integral in (4) . We have:

$\begin{aligned} \sum_{n=1}^\infty\frac{ 1}{n^4\binom{2n}{n}}&=2\int_0^{\pi/3}x\ln^{2}\left(2 \sin\left( \frac{x}{2}\right) \right)\,dx\\ &=8\int_0^{\pi/6}x\ln^{2}\left(2 \sin\left( x\right) \right)\,dx & \left(\frac{x}{2} \to x\right)\\ &=2\pi^2\int_0^{\pi/6}x\,dx-8\pi \int_0^{\pi/6}x^2\,dx+8\int_0^{\pi/6}x^3\,dx+16\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n}\int_0^{\pi/6}x\cos(2nx)\,dx & \left(\text{by eq.(5)} \right)\\ &=\frac{\pi^4}{36}-\frac{\pi^4}{81}+\frac{\pi^4}{648}+16\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n}\left(\frac{\pi}{12n}\sin\left(\frac{n \pi}{3} \right) -\frac{1}{2n}\int_0^{\pi/6}\sin(2nx)\,dx\right)\\ &=\frac{\pi^4}{36}-\frac{\pi^4}{81}+\frac{\pi^4}{648}+16\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}}{n}\left(\frac{\pi \sin\left(\frac{n \pi}{3} \right)}{12n} +\frac{\cos\left(\frac{n \pi}{3} \right)}{4n^2}-\frac{1}{4n^2}\right)\\ &=\frac{\pi^4}{36}-\frac{\pi^4}{81}+\frac{\pi^4}{648}+2\left(\frac{2\pi }{3}\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}\sin\left(\frac{n \pi}{3}\right)}{n^2} +2\sum_{n=1}^\infty \frac{\operatorname{H}_{n-1}\cos\left(\frac{n \pi}{3}\right)}{n^3}\right)-4\sum_{n=1}^\infty \frac{\operatorname{H}_{n}}{n^3}+4\sum_{n=1}^\infty \frac{1}{n^4} & (\text{by eq.(12)}) \\ &=\frac{15}{6}\zeta(4)-\frac{10}{9}\zeta(4)+\frac{5}{36}\zeta(4)-\frac{\zeta(4)}{18}-4\left( \frac{5}{4}\zeta(4)\right)+4\zeta(4) \\ &=\frac{15}{6}\zeta(4)-\frac{10}{9}\zeta(4)+\frac{5}{36}\zeta(4)-\frac{\zeta(4)}{18}-\zeta(4) \\ &=\frac{17}{36}\zeta(4) \qquad \blacksquare \end{aligned}$

Therefore, after rearranging terms we have

$\zeta(4)=\frac{36}{17}\sum_{n=1}^\infty\frac{ 1}{n^4\binom{2n}{n}}$

We used used the fact that (see here)

$\sum_{n=1}^\infty\frac{\operatorname{H}_n}{n^3}=\frac{5}{4}\zeta(4)$

References

J. M. Borwein, D. J. Broadhurst, J. Kamnitzer:Central Binomial Sums, Multiple Clausen Values and Zeta Values (arXiv:hep-th/0004153)

POLYLOGARITHMS, MULTIPLE ZETA VALUES, AND THE SERIES OF HJORTNAES AND COMTET Notes by Tim Jameson (December 2009)

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