7

I have encountered this integral and I am stuck evaluating it:$I=\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx$

My try was to expand the numerator into power series, indeed: $$x\cos^2x=\frac{x}{2}(1+\cos(2x)) =\frac{x}{2} +\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n-1} x^{2n+1}}{(2n)!}$$ And using $\zeta{(z)} \Gamma{(z)} =\int_0^{\infty} \frac{x^{z-1}}{e^x-1}dx$ gives: $$I=\frac{1}{2}\zeta{(2)}+ \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n-1}}{(2n)!} \zeta{(2n+2)} \Gamma{(2n+2)}=\frac{1}{2}\zeta{(2)}+ 2 \sum_{n=0}^{\infty} (-1)^n \zeta{(2n+2)}$$ Is there a way to simplify this? Or maybe another approach to this integral?

Edit: According to the answer in the comment, would this show that $\sum_{n=1}^{\infty} (-1)^{n-1} \zeta{(2n)}=\frac{\pi^2}{6}(2-3\text{ csch}^2(2\pi))+\frac{1}{16} $ ?

Zacky
  • 30,116
  • $\frac{1}{16}+\frac{\pi ^2}{12}-\frac{1}{4} \pi ^2 \text{csch}^2(2 \pi )$ – Mariusz Iwaniuk Apr 10 '18 at 10:41
  • See mickep answer. – Mariusz Iwaniuk Apr 10 '18 at 12:41
  • 1
    $$\begin{align}S&=\sum_{n=1}^{\infty}(-1)^{n-1} \zeta{(2n)}=\sum_{n=1}^{\infty}\left[\zeta(4n-2)-\zeta(4n)\right]\&=\sum_{n=1}^{\infty}\sum_{\color{red}{m=1}}^{\infty}\left[\frac1{m^{4n-2}}-\frac1{m^{4n}}\right]=\sum_{n=1}^{\infty}\sum_{\color{red}{m=2}}^{\infty}\left[\frac1{m^{4n-2}}-\frac1{m^{4n}}\right]\&=\sum_{\color{red}{m=2}}^{\infty}\sum_{\color{blue}{n=1}}^{\infty}\left[\frac1{m^{4n-2}}-\frac1{m^{4n}}\right]=-\color{red}{\sum_{m=2}^{\infty}}\color{blue}{\sum_{n=1}^{\infty}}\left(-m^2\right)^{-n}\&=\sum_{m=2}^{\infty}\frac1{m^2+1}=\frac{\pi}{2}\coth(\pi)-1\end{align}$$ – Hazem Orabi Jul 26 '18 at 11:33

4 Answers4

2

Try to use the following expansion: $$ \frac{1}{e^x-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{k=1}^{+\infty}e^{-kx}. $$ Then switch order of summation and integration, integrate, and you will end up with the series $$ \sum_{k=1}^{+\infty}\biggl(\frac{1}{2k^2}-\frac{4}{(4+k^2)^2}+\frac{1}{2(4+k^2)}\biggr). $$ I'm sure you can handle it.

mickep
  • 20,102
1

Hint. Another way is to use the residue theorem. Since, for a suitable function, we have $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{Residues of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} $$ we get $$\sum_{n\geq1}\frac{1}{2n^{2}}+\sum_{n\geq1}\left(\frac{1}{2\left(4+n^{2}\right)}-\frac{4}{\left(4+n^{2}\right)^{2}}\right)$$ $$=\frac{\pi^{2}}{12}+\frac{1}{16}+\frac{1}{2}\sum_{n\in\mathbb{Z}}\left(\frac{1}{2\left(4+n^{2}\right)}-\frac{4}{\left(4+n^{2}\right)^{2}}\right).$$ Now you have just to check that $f\left(z\right)=\frac{1}{2\left(4+z^{2}\right)}-\frac{4}{\left(4+z^{2}\right)^{2}}$ is an admissible function and $$\mathrm{Res}_{z=2i}\pi\cot\left(\pi z\right)f\left(z\right)=\mathrm{Res}_{z=-2i}\pi\cot\left(\pi z\right)f\left(z\right)=\frac{1}{4}\pi^{2}\mathrm{csch}^{2}\left(2\pi\right).$$

Community
  • 1
Marco Cantarini
  • 33,458
1

Well, I'd like to sum up everything because (as for me) the solution is here, in the answers of mickep and Marco Cantarini, but it's not rigorous for someone else.

So we have $$I=\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx=\frac12\int_0^{\infty} \frac{x}{e^x-1}dx+\frac12\int_0^{\infty}\frac{x\cos{2x}}{e^x-1}dx$$

As we know $\zeta{(z)} \Gamma{(z)} =\int_0^{\infty} \frac{x^{z-1}}{e^x-1}dx$ so the first integral equals to $\zeta{(2)} \Gamma{(2)} =\frac{\pi^2}6$ and the first summand is $\frac{\pi^2}{12}$.

The second summand could be expressed in the following way $$\frac12\int_0^{\infty}\frac{x\cos{2x}}{e^x-1}dx=\frac12\int_0^{\infty}\left(x\cos{2x}\sum_{n=1}^{\infty}e^{-nx}\right) dx=$$ $$=\frac12\sum_{n=1}^{\infty}\int_0^{\infty}e^{-nx}x\cos{2x}dx$$ To find the latest integral one can use Laplace transform of function $x\cos{2x}$. According to the table of Laplace transforms $$\mathfrak{L}\{ x \cos{2x}\}(s)=\int_0^{\infty}e^{-sx}x\cos{2x}dx=\frac{s^2-4}{(s^2+4)^2}$$ The integral in the middle is required integral when $s=n$.

Thus the second summand becomes $$\frac12\sum_{n=1}^{\infty}\frac{n^2-4}{(n^2+4)^2}=\frac12\left(\frac12\sum_{n=-\infty}^{\infty}\frac{n^2-4}{(n^2+4)^2}+\frac18\right)=$$ $$= \frac1{16}-\frac14\left(\mathrm{Res}_{z=2i}\left(\pi\cot\left(\pi z\right)\frac{z^2-4}{(z^2+4)^2}\right)+\mathrm{Res}_{z=-2i}\left(\pi\cot\left(\pi z\right)\frac{z^2-4}{(z^2+4)^2}\right)\right)=$$ $$=\frac1{16}-\frac{1}{4}\pi^{2}\mathrm{csch}^{2}\left(2\pi\right)$$

And finally we have $I=\frac{\pi^2}{12}+\frac1{16}-\frac{1}{4}\pi^{2}\mathrm{csch}^{2}\left(2\pi\right)$.

Mikalai Parshutsich
  • 620
1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}{x\cos^{2}\pars{x} \over \expo{x} - 1}\,\dd x}} = {1 \over 2}\int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x + {1 \over 2}\int_{0}^{\infty}{x\cos\pars{2x} \over \expo{x} - 1}\,\dd x \\[5mm] &\ \left\{\begin{array}{rcl} \ds{{1 \over 2}\int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x} & \ds{=} & \ds{{1 \over 2}\int_{0}^{\infty}{x\expo{-x} \over 1 - \expo{-x}}\,\dd x = {1 \over 2}\sum_{n = 0}^{\infty}\int_{0}^{\infty}x\expo{-\pars{n + 1}x}\dd x} \\ & \ds{=} & \ds{{1 \over 2}\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}^{\, 2}} = {1 \over 2}\,{\pi^{2} \over 6} = \bbx{\pi^{2} \over 12}} \\[1cm] \ds{{1 \over 2}\int_{0}^{\infty}{x\cos\pars{2x} \over \expo{x} - 1}\,\dd x} & \ds{=} & \ds{\left.{1 \over 2}\,\totald{}{a}\int_{0}^{\infty}{\sin\pars{ax} \over \expo{x} - 1}\,\dd x\,\right\vert_{\ a\ =\ 2}} \\[2mm] & \ds{=} & \ds{{1 \over 2}\,\totald{}{a}\bracks{\pi a\coth\pars{\pi a} - 1 \over 2a}_{\ a\ =\ 2}} \\[2mm] & \ds{=} & \ds{{1 \over 2}\,\totald{}{a}\bracks{{1 \over 2a^{2}} - {\pi^{2} \over 2}\,\mrm{csch}^{2}\pars{\pi a}}_{\ a\ =\ 2}} \\[2mm] & \ds{=} & \bbx{\ds{{1 \over 16} - {\pi^{2} \over 4}\,\mrm{csch}^{2}\pars{2\pi}}} \end{array}\right. \end{align} Then, $$ \bbx{\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}{x\cos^{2}\pars{x} \over \expo{x} - 1}\,\dd x}} = {\pi^{2} \over 12} + {1 \over 16} - {\pi^{2} \over 4}\,\mrm{csch}^{2}\pars{2\pi}} \approx 0.8849 $$

The integral $\ds{\int_{0}^{\infty}{\sin\pars{ax} \over \expo{x} - 1}\,\dd x = {\pi a\coth\pars{\pi a} - 1 \over 2a}}$ has already been evaluated in this link.

Felix Marin
  • 94,079