# Closed form for ${\large\int}_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$

I'm interested in a closed form for this simple looking integral: $$I=\int_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$$ Numerically, $$I\approx0.235708612100161734103782517656481953570915076546754616988...$$ Note that if we try to split the integral into two parts, each with only one term in the numerator, then both parts will be divergent.

• Where did you come across this? – Aryabhata Jan 11 '15 at 18:57
• It may be useful to know that $$\int\limits_0^\infty{\frac{{x-\sin x}}{{x\left({{e^x}-1}\right)}}}dx =\gamma+{\mathop{\rm Im}\nolimits}\left({\ln\Gamma\left({1+i}\right)}\right)$$ – user111187 Jan 11 '15 at 19:05
• The final answer comes from a community effort (many thanks to Vladimir Reshetnikov and robjohn): $$\color{purple}{\int_{0}^{+\infty}\frac{x-\sin x}{x^2(e^x-1)}\,dx=\frac{1}{2}+\frac{5\pi}{24}-\log\sqrt{2\pi}+\frac{1}{4\pi}\Li_2(e^{-2\pi})}.$$ – Jack D'Aurizio Jan 11 '15 at 20:08

$$\frac{x-\sin x}{x^2}=\sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n+1)!}x^{2n-1},$$ and since: $$\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\,dx = (2n-1)!\cdot \zeta(2n),$$ we have: $$\begin{eqnarray*} &&\int_{0}^{+\infty}\frac{x-\sin x}{x^2(e^x-1)}\,dx = \sum_{n\geq 1}\frac{(-1)^{n+1}}{2n(2n+1)}\zeta(2n)\\&=&\color{red}{\sum_{n\geq 1}\left(-1+n\arctan\frac{1}{n}+\frac{1}{2}\,\log\left(1+\frac{1}{n^2}\right)\right)}\\&=&\color{blue}{\log\sqrt{\frac{\sinh \pi}{\pi}}+\sum_{n\geq 1}\left(-1+n\arctan\frac{1}{n}\right)}.\tag{1}\end{eqnarray*}$$ Combining this identity with the robjonh's answer to another question, we finally get: $$\color{purple}{\int_{0}^{+\infty}\frac{x-\sin x}{x^2(e^x-1)}\,dx=\frac{1}{2}+\frac{5\pi}{24}-\log\sqrt{2\pi}+\frac{1}{4\pi}\operatorname{Li}_2(e^{-2\pi})}.\tag{2}$$

On the other hand, the identity claimed by user111187, $$\int_{0}^{+\infty}\frac{x-\sin x}{x(e^x-1)} = \gamma+\Im\log\Gamma(1+i)\tag{3}$$ follows from the integral representation for the $\log\Gamma$ function and for the Euler-Mascheroni constant. By considering the Weierstrass product for the $\Gamma$ function, $$\Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{\frac{z}{n}}$$ we have: $$\log\Gamma(z+1) = -\gamma z + \sum_{n\geq 1}\left(\frac{z}{n}-\log\left(1+\frac{z}{n}\right)\right)$$ so: $$\int_{0}^{+\infty}\frac{x-\sin x}{x(e^x-1)}\,dx = \sum_{n\geq 1}\left(\frac{1}{n}-\arctan\frac{1}{n}\right).\tag{4}$$

• It looks like the red sum can be simplified using robjohn's answer to another Laila's question, and using $\prod_{n\ge1}\left(1+n^{-2}\right)=\frac{\sinh\pi}{\pi}$. – Vladimir Reshetnikov Jan 11 '15 at 19:46
• @VladimirReshetnikov: I see, improved it. Thanks for the review process :D – Jack D'Aurizio Jan 11 '15 at 20:04
• Very nice answer. Thanks! – Laila Podlesny Jan 11 '15 at 21:15


\begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{x - \sin\pars{x} \over \pars{\expo{x} - 1}x^{2}}\,\dd x} =\int_{0}^{\infty}{1 \over \pars{\expo{x} - 1}x}\,{x - \sin\pars{x} \over x}\,\dd x \\[5mm]&=\int_{0}^{\infty}{1 \over \pars{\expo{x} - 1}x}\,\ \overbrace{\half\int_{-1}^{1}\pars{1 + \ic k x - \expo{\ic k x}}\,\dd k} ^{\dsc{x - \sin\pars{x} \over x}}\ \,\dd x\ =\ \half\int_{-1}^{1}\int_{0}^{\infty}{1 + \ic k x - \expo{\ic k x} \over \pars{\expo{x} - 1}x}\,\dd x\,\dd k \\[5mm]&=\half\int_{-1}^{1}\int_{0}^{\infty} {\expo{-x} + \ic k x\expo{-x} - \expo{-\pars{1 - \ic k}x} \over 1 - \expo{-x}}\ \overbrace{\int_{0}^{\infty}\expo{-xt}\,\dd t}^{\dsc{1 \over x}}\ \dd x\,\dd k \\[5mm]&=\half \int_{-1}^{1}\int_{0}^{\infty}\sum_{n\ =\ 0}^{\infty}\ \int_{0}^{\infty} \bracks{\expo{-\pars{n + 1 + t}x}\ +\ \ic k x\expo{-\pars{n + 1 + t}x}\ -\ \expo{-\pars{n + 1 + t - \ic k}x}\,\,} \,\dd x\,\dd t\,\dd k \\[5mm]&=\half \int_{-1}^{1}\int_{0}^{\infty}\sum_{n\ =\ 0}^{\infty}\ \bracks{{1 \over n + 1 + t}\ +\ {\ic k \over \pars{n + 1 + t}^{2}}\ -\ {1 \over n + 1 + t - \ic k}}\,\dd t\,\dd k \\[5mm]&=\half \int_{-1}^{1}\int_{0}^{\infty}\sum_{n\ =\ 0}^{\infty}\ \bracks{{\ic k \over \pars{n + 1 + t}^{2}} -{\ic k \over \pars{n + 1 + t}\pars{n + 1 + t - \ic k}}}\,\dd t\,\dd k \\[5mm]&=\half\int_{-1}^{1}\int_{0}^{\infty} \bracks{\Psi\pars{1 + t -\ic k} - \Psi\pars{1 + t}}\,\dd t\,\dd k \end{align} where $\ds{\Psi}$ is the Digamma Function.
Then, \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{x - \sin\pars{x} \over \pars{\expo{x} - 1}x^{2}}\,\dd x} =\left.\half\int_{-1}^{1} \ln\pars{\Gamma\pars{1 + t - \ic k} \over \Gamma\pars{1 + t}} \right\vert_{\, t\ =\ 0}^{\, t\ \to\ \infty}\,\dd k \\[5mm]&=-\,\half\int_{-1}^{1}\ln\pars{\Gamma\pars{1 - \ic k}}\,\dd k =\dsc{-\,\Re\int_{0}^{1}\ln\pars{\Gamma\pars{1 - \ic k}}\,\dd k}\tag{1} \end{align}
However, \begin{align}&\dsc{-\,\Re\int_{0}^{1}\ln\pars{\Gamma\pars{1 - \ic k}}\,\dd k} =-\,\Re\int_{0}^{1}\ln\pars{-\ic k\Gamma\pars{-\ic k}}\,\dd k \\[5mm]&=-\int_{0}^{1}\ln\pars{k}\,\dd k -\Re\int_{0}^{1}\ln\pars{\Gamma\pars{-\ic k}}\,\dd k =1-\Re\int_{0}^{1} \ln\pars{\pi \over \Gamma\pars{1 + \ic k}\sin\pars{\pi\bracks{-\ic k}}}\,\dd k \\[5mm]&=1 - \ln\pars{\pi} + \Re\int_{0}^{1}\ln\pars{ \Gamma\pars{1 + \ic k}}\,\dd k +\Re\int_{0}^{1}\ln\pars{-\ic\sinh\pars{\pi k}}\,\dd k \\[5mm]&=1 - \ln\pars{\pi} +\dsc{\Re\int_{0}^{1}\ln\pars{ \Gamma\pars{1 - \ic k}}\,\dd k} +\int_{0}^{1}\ln\pars{\sinh\pars{\pi k}}\,\dd k \\[1cm]&\imp\quad \dsc{-\,\Re\int_{0}^{1}\ln\pars{\Gamma\pars{1 - \ic k}}\,\dd k} =\half - \half\,\ln\pars{\pi} +\half\int_{0}^{1}\ln\pars{\sinh\pars{\pi k}}\,\dd k \\[5mm]&=\half - \half\,\ln\pars{\pi} +\half\int_{0}^{1}\bracks{\pi k + \ln\pars{1 - \expo{-2\pi k}} - \ln\pars{2}} \,\dd k \\[5mm]&=\half - \half\,\ln\pars{2\pi} + {1 \over 4}\,\pi -\half\sum_{n\ =\ 1}{1 \over n}\int_{0}^{1}\expo{-2n\pi k}\,\dd k \\[5mm]&=\half - \half\,\ln\pars{2\pi} + {1 \over 4}\,\pi -\half\sum_{n\ =\ 1}{1 \over n}{\expo{-2n\pi} - 1 \over -2n\pi} \\[5mm]&=\half - \half\,\ln\pars{2\pi} + {1 \over 4}\,\pi +{1 \over 4\pi}\ \overbrace{\sum_{n\ =\ 1}{\pars{\expo{-2\pi}}^{n} \over n^{2}}} ^{\dsc{\Li{2}\pars{\expo{-2\pi}}}}\ -{1 \over 4\pi}\, \overbrace{\sum_{n\ =\ 1}{1 \over n^{2}}}^{\dsc{\pi^{2} \over 6}} \\[5mm]&=\half - \half\,\ln\pars{2\pi} + {5 \over 24}\,\pi +{1 \over 4\pi}\,\Li{2}\pars{\expo{-2\pi}} \end{align}
Replacing in expression $\pars{1}$: \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{x - \sin\pars{x} \over \pars{\expo{x} - 1}x^{2}}\,\dd x} =\color{#66f}{\large\half - \half\,\ln\pars{2\pi} + {5 \over 24}\,\pi +{1 \over 4\pi}\,\Li{2}\pars{\expo{-2\pi}}} \\[5mm]&\approx {\tt 0.2357} \end{align}

This is another solution just for reference.

The laplace transform of $\displaystyle\frac{x-\sin{x}}{x^2}$ is given by \begin{align} \mathcal{L}_s\left(\frac{x-\sin{x}}{x^2}\right) &=\int^s_\infty\int^t_\infty\frac{1}{u^2}-\frac{1}{1+u^2}\ du\ dt\tag1\\ &=\int^s_\infty-\frac{1}{t}-\arctan{t}+\frac{\pi}{2}\ dt\\ &=-\ln{s}+\frac{1}{2}\ln(1+s^2)-s\arctan{s}+\frac{\pi}{2}s-1\\ &=\frac{1}{2}\ln\left(1+\frac{1}{s^2}\right)+s\arctan\left(\frac{1}{s}\right)-1 \end{align} Thus \begin{align} \int^\infty_0\frac{x-\sin{x}}{x^2(e^x-1)}dx &=\sum^\infty_{n=1}\mathcal{L}_n\left(\frac{x-\sin{x}}{x^2}\right)\tag2\\ &=\frac{1}{2}\sum^\infty_{n=1}\ln\left(1+\frac{1}{n^2}\right)+\sum^\infty_{n=1}\left(n\arctan\left(\frac{1}{n}\right)-1\right)\\ &=\frac{1}{2}\ln\left.\frac{\pi z}{\pi}\prod^\infty_{n=1}\left(1+\frac{z^2}{n^2}\right)\right|_{z=1}+\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^k}{(2k+1)n^{2k}}\tag{3}\\ &=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\sum^\infty_{k=1}\frac{(-1)^k\zeta(2k)}{(2k+1)}\tag4\\ &=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2i}\int^{i}_0\left(1-\pi z\cot(\pi z)\right)\ dz\tag5\\ &=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2}-\frac{1}{8\pi}\int^{\exp(-2\pi)}_1\frac{\ln{u}(1+u)}{u(1-u)}du\tag6\\ &=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2}-\frac{1}{8\pi}\left[2\mathrm{Li}_2(1-u)+\frac{\ln^2{u}}{2}\right]^{\exp(-2\pi)}_1\tag7\\ &=\color{#E2062C}{\frac{1}{2}-\frac{\pi}{4}+\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)-\frac{1}{4\pi}\mathrm{Li}_2\left(1-e^{-2\pi}\right)} \end{align} Explanation:

$(1)$: Differentiated under the integral twice.
$(2)$: Expanded $(e^{x}-1)^{-1}$.
$(3)$: Expanded $\arctan\left(n^{-1}\right)$.
$(4)$: Recognised the Weierstrass product for $\sinh$, summed in $n$.
$(5)$: Used the fact that $\displaystyle\pi z\cot(\pi z)=1-2\sum^\infty_{k=1}\zeta(2k)z^{2k}$.
$(6)$: Substitued $u=e^{2\pi iz}$.
$(7)$: $\displaystyle \frac{\ln{u}(1+u)}{u(1-u)}=\frac{2\ln{u}}{1-u}+\frac{\ln{u}}{u}$.

Here is a bit of a generalization, and what I think an easier solution.

Start with $\displaystyle \int_0^{\infty} \sin(ax) \,e^{-xn}\,dx=\frac{a}{a^2+n^2}$. (Easily established through integration by parts.)

Integrate both sides w.r.t. $a$ from $0$ to $t$, to get $\displaystyle \int_0^{\infty} \frac{1-\cos(tx)}{x} \,e^{-xn}\,dx=\frac12 \ln\left(1+\frac{t^2}{n^2}\right).$

Sum both sides from $n=1$ to $\infty$, to get $\displaystyle \int_0^{\infty} \frac{1-\cos(tx)}{x(e^x-1)}\,dx=\frac12 \ln \dfrac{\sinh(\pi t)}{\pi t}$

Finally, integrate both sides w.r.t. $t$ from $0$ to $a$, to get: $$\int_0^{\infty} \frac{xa-\sin(xa)}{x^2(e^x-1)}\,dx=\frac12 \int_0^a \ln \frac{\sinh(\pi t)}{\pi t}\,dt \\\\=-\frac12 a \ln\pi-\frac12(a\ln a-a)+\frac12\int_0^a (\ln(1-e^{-2\pi t})-\ln2+\pi t)\,dt \\\\=-\frac12 a \ln(2\pi)-\frac12 a\ln a+\frac{a}{2}+\frac{\pi}{4}a^2-\frac12\int_0^a \sum_{n=1}^{\infty} \frac{e^{-2\pi t n}}{n}\,\,dt \\\\=\frac{\pi}{24}(6a^2-1)+\frac{a}{2}(1-\ln(2\pi))-\frac12 a\ln a+\frac1{4\pi}\operatorname{Li}_2(e^{-2\pi a})$$

In particular, when $a=1$, $$\int_0^{\infty} \frac{x-\sin(x)}{x^2(e^x-1)}\,dx=\frac12+\frac{5}{24}\pi-\frac12\ln(2\pi)+\frac1{4\pi}\operatorname{Li}_2(e^{-2\pi }).$$