오늘의 계산 56 - An infinite summation involving binomial coefficient

Problem 1. Prove that[각주:1]

$$\begin{array}{}\text{(1)}& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4(\genfrac{}{}{0ex}{}{2n}{n})}=\frac{17{\pi }^4}{3240}.\end{array}$$

 

Proof. We divide the proof into several steps.

1. Reduction to an integral representation

Let $S$ denote the summation in question. By using the Lemma 1 in Today's Calculation 29, we can represent $S$ as an integral. Then by the successive application of integration by parts, we obtain

$$\begin{array}{rl}S& =8{\int }_0^1\frac{1}{x}{\int }_0^{x/2}\frac{{\arcsin }^{2}t}{t}\mathrm{d}t\mathrm{d}x=-8{\int }_0^1\frac{{\arcsin }^2(x/2)}{x}\log x\mathrm{d}x\\ & =8{\int }_0^1\frac{\arcsin (x/2)}{\sqrt{1-(x/2{)}^2}}{\log }^{2}x\frac{\mathrm{d}x}{2}.\end{array}$$

Thus with the substitution $x=2\sin \theta$, we have

$$S=8{\int }_0^{\frac{\pi }{6}}\theta {\log }^2(2\sin \theta )\mathrm{d}\theta .$$

To evaluate this integral, note that for $0<\theta <\frac{\pi }{6}$ we have

$$e^{i\theta }\cdot 2\sin \theta =i\cdot (1-e^{2i\theta }).$$

Taking logarithm (with the branch cut $(-\mathrm{\infty },0]$ as usual) to both sides, it follows that

$$i\theta +\log (2\sin \theta )=\frac{i\pi }{2}+\log (1-e^{2i\theta }).$$

Cubing both sides and integrating on $(0,\frac{\pi }{6})$ and taking imaginary parts only,

$$\begin{array}{}\text{(2)}& S=\frac{2}{3}{\left(\frac{\pi }{6}\right)}^4+\frac{8}{3}\mathrm{\Im }{\int }_0^{\frac{\pi }{6}}{(\frac{i\pi }{2}+\log (1-e^{2i\theta }))}^3\mathrm{d}\theta .\end{array}$$

2. Some complex-analysis techniques

Now we focus on the integral in the imaginary part:

$$\begin{array}{}\text{(3)}& I:={\int }_0^{\frac{\pi }{6}}{(\frac{i\pi }{2}+\log (1-e^{2i\theta }))}^3\mathrm{d}\theta .\end{array}$$

Once we evaluate the imaginary part of $I$, the identity $\text{(2)}$ immediately gives us the answer. We first make the substitution $z=1-e^{2i\theta }$ and $\omega =e^{-i\pi /3}$ to obtain

$$I={\int }_0^{\omega }{(\frac{i\pi }{2}+\log z)}^3\frac{\mathrm{d}z}{2i(z-1)}.$$

Here, the path of integration is a circular arc joining from $0$ to $\omega$ centered at $1$.

But since the integrand is analytic for $0<\mathrm{\Re }z<1$, we may change the path of integration as $z=\omega t$ for $0\le t\le 1$. This gives

$$I=\frac{1}{2i}{\int }_0^1{(\frac{i\pi }{6}+\log t)}^3\frac{\omega \mathrm{d}t}{\omega t-1}.$$

Plugging $t=e^{-x}$, $I$ reduces to

$$\begin{array}{rl}I& =\frac{1}{2}{\int }_0^{\mathrm{\infty }}{(\frac{\pi }{6}+ix)}^3\frac{\omega e^{-x}}{1-\omega e^{-x}}\mathrm{d}x=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}{\omega }^n{\int }_0^{\mathrm{\infty }}{(\frac{\pi }{6}+ix)}^3{e}^{-nx}\mathrm{d}x\\ & =\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}{\omega }^n(-\frac{6i}{n^4}-\frac{\pi }{n^3}+\frac{i{\pi }^2}{12n^2}+\frac{{\pi }^3}{216n}).\end{array}$$

Taking the imaginary part,

$$\begin{array}{}\text{(4)}& \begin{array}{rl}\mathrm{\Im }I& =-3\sum _{n=1}^{\mathrm{\infty }}\frac{\cos (n\pi /3)}{n^4}+\frac{\pi }{2}\sum _{n=1}^{\mathrm{\infty }}\frac{\sin (n\pi /3)}{n^3}\\ & +\frac{{\pi }^2}{24}\sum _{n=1}^{\mathrm{\infty }}\frac{\cos (n\pi /3)}{n^2}-\frac{{\pi }^3}{432}\sum _{n=1}^{\mathrm{\infty }}\frac{\sin (n\pi /3)}{n}.\end{array}\end{array}$$

3. Evaluation of a series[각주:2]

Note that for $0<\theta <\pi$, we have

$$\sum _{n=1}^{\mathrm{\infty }}\frac{\sin n\theta }{n}=\mathrm{\Im }\sum _{n=1}^{\mathrm{\infty }}\frac{e^{i\theta }}{n}=-\mathrm{\Im }\log (1-e^{i\theta })=\frac{\pi -\theta }{2}.$$

Integrating both sides, we obtain

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1-\cos n\theta }{n^2}=\frac{\theta (2\pi -\theta )}{4}⟹\sum _{n=1}^{\mathrm{\infty }}\frac{\cos n\theta }{n^2}=\frac{{\theta }^2}{4}-\frac{\pi \theta }{2}+\frac{{\pi }^2}{6}.$$

Repeating this procedure, we obtain

$$\sum _{n=1}^{\mathrm{\infty }}\frac{\sin n\theta }{n^3}=\frac{{\theta }^3}{12}-\frac{\pi {\theta }^2}{4}+\frac{{\pi }^2\theta }{6}$$

and

$$\sum _{n=1}^{\mathrm{\infty }}\frac{\cos n\theta }{n^4}=-\frac{{\theta }^4}{48}+\frac{\pi {\theta }^3}{12}-\frac{{\pi }^2{\theta }^2}{12}+\frac{{\pi }^4}{90}.$$

Plugging $\theta =\frac{\pi }{3}$, we have

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{\sin (n\pi /3)}{n^2}& =\frac{\pi }{3}\\ \sum _{n=1}^{\mathrm{\infty }}\frac{\cos (n\pi /3)}{n^2}& =\frac{{\pi }^3}{36}\\ \sum _{n=1}^{\mathrm{\infty }}\frac{\sin (n\pi /3)}{n^3}& =\frac{5{\pi }^3}{162}\\ \sum _{n=1}^{\mathrm{\infty }}\frac{\cos (n\pi /3)}{n^4}& =\frac{91{\pi }^4}{19440}\end{array}$$

Plugging these to $\text{(4)}$, we have

$$\mathrm{\Im }I=\frac{23{\pi }^4}{12960}⟹S=\frac{17{\pi }^4}{3240}=\frac{17}{36}\zeta (4)$$

as desired.

각주

1. Integral, Proof of ... - Math StackExchange [본문으로]
2. This is closely related to the Bernoulli polymomials. See here for its connection. [본문으로]