I encountered the following integral and numerical approximations tentatively suggest that it might have a simple closed form:

$${\large\int}_0^\infty\left[\frac1{x^4}-\frac1{2x^3}+\frac1{12\,x^2}-\frac1{\left(e^x-1\right)x^3}\right]dx\stackrel{\color{gray}?}=\frac{\zeta(3)}{8\pi^2}\tag{$\diamond$}$$ (Update: I fixed a typo: replaced $4\pi^2$ with $8\pi^2$ in the denominator)

I have only about $800$ decimal digits that agree with the conjectured value, calculated using Mathematica. Unfortunately, its numerical algorithms become unstable when I try to increase precision. Maple refuses to numerically evaluate this integral altogether.

Obviously, the first three terms of the integrand have elementary antiderivatives, but I was not able to find a closed-form antiderivative (either elementary or using known special functions) for the last one.

I'm asking for your help in proving (or disproving) the $(\diamond)$.

  • 3
    $\begingroup$ An observation: This resembles the integral rep. of the riemann zeta function togehter combined with their analytic continunation to negative numbers. $\endgroup$
    – tired
    Commented Jan 10, 2016 at 21:19
  • $\begingroup$ Contour integration where you integrate from minus to plus infinity and close the contour in the upper half plane. The odd functions in the integrand should be modified, using the symmetry of the 1/(exp(x) - 1) function. $\endgroup$ Commented Jan 10, 2016 at 21:21
  • 8
    $\begingroup$ Your integral is nothing more than the rigorous way of asking for the regularization of $~\displaystyle\int_0^\infty\frac{x^n}{e^x-1}~dx~=~n!~\zeta(n+1)~$ to $n<0,$ or, in other words, for evaluating $\lim\limits_{n\to-3}n!~\zeta(n+1).$ $\endgroup$
    – Lucian
    Commented Jan 11, 2016 at 7:27
  • 3
    $\begingroup$ In general, $\lim\limits_{n\to-(2k+1)}~n!~\zeta(n+1) ~=~ (-1)^k\cdot\pi\cdot(2\pi)^n\cdot\zeta(-n).$ $\endgroup$
    – Lucian
    Commented Jan 11, 2016 at 7:47

4 Answers 4


What about the Laplace transform? By using it, we have that our integral equals:

$$ I= \frac{1}{36}\int_{0}^{+\infty}\left(1-3 s+6 s^2-6 s^3 \psi'(1+s)\right)\,ds$$ and in this form Mathematica is perfectly able to state that $I=\frac{\zeta(3)}{\color{red}{8}\pi^2}$.

I just used: $$\mathcal{L}^{-1}\left(\frac{1}{x^4}\right)=\frac{s^3}{6},\qquad \mathcal{L}\left(1-\frac{x}{2}+\frac{x^2}{12}-\frac{x}{e^x-1}\right)=\frac{1-3 s+6 s^2}{6 s^3}-\psi'(1+s) $$ together with: $$ \int_{0}^{+\infty}f(x)g(x)\,dx = \int_{0}^{+\infty}(\mathcal{L} f)(s)(\mathcal{L}^{-1}g)(s)\,ds.$$

  • $\begingroup$ Thanks, I fixed a typo. Could you provide some reference about this use of the Laplace transform? I'm only familiar with its application to certain differential equations. $\endgroup$ Commented Jan 10, 2016 at 21:39
  • $\begingroup$ shouldn't it be $$ \int_{0}^{+\infty}f(x)g(x)^*\,dx = \int_{0}^{+\infty}(\mathcal{L} f)(2 i \pi s)(\mathcal{L} g)(2 i \pi s)^*\,ds.$$ where $z^*$denotes the complex conjugate ? or this other formula on the real axis : $$\int_0^\infty \frac{f(t)}{t}\, dt = \int_0^\infty (\mathcal{L} f)(s)\, ds$$ $\endgroup$
    – reuns
    Commented Jan 10, 2016 at 23:40
  • $\begingroup$ @user1952009: your second formula is consistent with mine, since $\mathcal{L}^{-1}(1/x)=1$. $\endgroup$ Commented Jan 11, 2016 at 0:00
  • $\begingroup$ @Jack D'Aurizio where does your formula comes from ? in my opinion it is true because we assume $F(s)G(s) \to 0$ when $|s| \to \infty$ so that we can change the contour of my 1st formula $\endgroup$
    – reuns
    Commented Jan 11, 2016 at 0:09
  • 2
    $\begingroup$ ok I'm finally convinced. for real vectors $a,b$ : $\langle a, b \rangle = b^t a$ so that for any inversible matrix $M$ : $$\langle M a, (M^{-1})^tb \rangle = b^t M^{-1} M a = \langle a, b \rangle$$ which is also true for (bounded ?) linear operators in Hilbert space, which is the Laplace transform on the real axis. thank you. $\endgroup$
    – reuns
    Commented Jan 11, 2016 at 11:18

$$\text{for} \ \ \Re(s) > 1 : \qquad \int_0^\infty \frac{x^{s-1}}{e^x-1}dx = \Gamma(s) \zeta(s)$$ $\Gamma(s) \zeta(s)$ is meromorphic so that we can easily jump other its poles which are at $1,0,-2n+1$ for $n \in \mathbb{N}^*$.

the pole at $s=1$ is of residue $1$, so that :

$$\text{for} \ \ \Re(s) \in ]0;1[ : \qquad \int_0^\infty \frac{x^{s-1}}{e^x-1} - x^{s-2} dx = \Gamma(s) \zeta(s)$$

the pole at $s=0$ is of residue $\zeta(0) = -1/2$, so that :

$$\text{for} \ \ \Re(s) \in ]-1;0[ : \qquad \int_0^\infty \frac{x^{s-1}}{e^x-1} - x^{s-2} +\frac{x^{s-1}}{2} dx = \Gamma(s) \zeta(s)$$

the pole at $s=-1$ is of residue $-\zeta(-1) = 1/12$, so that : $$\text{for} \ \ \Re(s) \in ]-3;-1[ : \qquad \int_0^\infty \frac{x^{s-1}}{e^x-1} - x^{s-2} + \frac{x^{s-1}}{2} - \frac{x^{s}}{12} dx = \Gamma(s) \zeta(s)$$

and finally when $s \to 0$ : $\Gamma(s-2) \approx \frac{1}{2s}$ and $\zeta(s-2) \approx s \zeta^{\prime}(-2) = -s\frac {2} {2 (2\pi)^{2}} \zeta (3)$ so that your integral is $$\lim_{s\to 0} - \Gamma(s-2) \zeta(s-2) = \frac { \zeta (3)} {8 \pi ^2}$$




  • $\begingroup$ which contour u are using exactly? $\endgroup$
    – tired
    Commented Jan 11, 2016 at 0:11
  • $\begingroup$ only real integrals. just prove the 1st pole jumping to be convinced. it relies on the fact that $\frac{1}{e^x-1} - \frac{1}{x}$ is $\mathcal{O}(1)$ when $x \to 0$. and the fact that the only pole of $\Gamma(s) \zeta(s)$ for $\Re(s) > 0$ is the one at $s=1$ of order $1$ and residue $1$ tells us exactly the same thing. $\endgroup$
    – reuns
    Commented Jan 11, 2016 at 0:16

From the summation identity of zeta function: $$ \boxed{ {\,}\\ \quad \color{Blue}{\sum_{n=0}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)}{(n+1)!}=0 \qquad\colon\space Re\{s\}\lt1} \quad \\{\,} } $$ $$ \begin{align} \Gamma(s)\zeta(s) &= -\sum_{n=1}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)}{(n+1)!} = -\int_{0}^{\infty}\frac{x^{s-2}}{e^x-1}\left(\sum_{n=1}^{\infty}\frac{x^{n+1}}{(n+1)!}\right)\,dx \\[2mm] &= -\int_{0}^{\infty}\frac{x^{s-2}}{e^x-1}\left(e^x-1+x\right)\,dx = \int_{0}^{\infty}x^{s-2}\left(\frac{x}{e^x-1}-1\right)\,dx \\[4mm] \Gamma(s-1)\zeta(s-1) &= -\frac{\Gamma(s)\zeta(s)}{2!}-\sum_{n=2}^{\infty}\frac{\Gamma(n+s)\zeta(n+s)}{(n+1)!} \\[2mm] &= \int_{0}^{\infty}x^{s-3}\left(\frac{x}{e^x-1}-1+\frac{x}{2}\right)\,dx \qquad\cdots\,\implies \end{align} $$

$$ \color{blue}{\Gamma(s-N)\zeta(s-N)=\int_{0}^{\infty}x^{s-N-2}\left[\frac{x}{e^x-1}-\left(\sum_{n=0}^{N}B_{n}\frac{x^n}{n!}\right)\right]\,dx} $$

$$ {\small \,0\lt\,Re\{s\}\,\lt1 ,\quad N\in\{\,0,\,1,\,2,\,\cdots\,\} ,\quad B_{n}\,\,{Bernoulli\,Number} ,\quad B_{1}=-1/2} $$

$$ \begin{align} \color{red}{I} &= \int_{0}^{\infty}\left[\frac{1}{x^4}-\frac{1}{2\,x^3}+\frac{1}{12\,x^2}-\frac{1}{\left(e^x-1\right)\,x^3}\right]\,dx \\[3mm] &= -\int_{0}^{\infty}x^{-4}\left[\frac{x}{e^x-1}-1+\frac{x}{2}-\frac{x^2}{12}\right]\,dx \\[3mm] &= -\int_{0}^{\infty}x^{\color{red}{0-2}-2}\left[\frac{x}{e^x-1}-\left(1\frac{x^0}{0!}-\frac{1}{2}\frac{x^1}{1!}+\frac{1}{6}\frac{x^2}{2!}\right)\right]\,dx \\[3mm] &= -\lim_{s\rightarrow0}\Gamma(s-2)\zeta(s-2)=-\frac{\zeta'(-2)}{2}=\color{red}{\frac{\zeta(3)}{8\pi^2}} \end{align} $$

  • $\begingroup$ Nice answer....... +1 $\endgroup$
    – TShiong
    Commented Feb 17, 2023 at 7:16

Utilize the known integral $$\frac1\pi\int_0^\infty \frac{\sin\frac{xt}{2\pi}}{e^t-1}dt = \frac12 \coth\frac x2-\frac1x$$ to evaluate \begin{align} &\int_0^\infty\left[\frac1{x^4}-\frac1{2x^3}+\frac1{12\,x^2}-\frac1{\left(e^x-1\right)x^3}\right]dx\\ =&\int_0^\infty\bigg(\frac1{x^4}+\frac1{12\,x^2}-\frac{\coth\frac x2}{2x^3}\bigg)dx\\ =&\int_0^\infty\bigg(\frac x{12}-\frac1{\pi}\int_0^\infty \frac{\sin\frac{xt}{2\pi}}{e^t-1}dt\bigg)\frac{dx}{x^3}\\ \overset{ibp}= &\int_0^\infty\bigg(\frac1{24}-\frac1{4\pi^2 }\int_0^\infty \frac{t\cos\frac{xt}{2\pi}}{e^t-1}dt\bigg) \frac{dx}{x^2}\\ \overset{ibp}=& \ \frac1{8\pi^3}\int_0^\infty\int_0^\infty \frac{t^2}{e^t-1} \frac{\sin\frac{xt}{2\pi}}{x} dx\ dt\\ =&\ \frac{1}{16\pi^2}\int_0^\infty \frac{t^2}{e^t-1}dt =\frac1{8\pi^2}\zeta(3) \end{align}


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