I would like to calculate the following integral:

$\displaystyle\int_{-\infty}^{+\infty}\frac{1}{\big(e^x-1-x\big)^2+4\pi^2} ~dx$

My attempts:

I have tried to use the complex analysis, in particular the Cauchy’s residue theorem which is a powerful tool to evaluate line integrals of analytic functions over closed curves, but the denominator

$\big(e^x-1-x\big)^2+4\pi^2=\big(e^x\!-\!1\!-\!x\!+\!2\pi i\big)\big(e^x\!-\!1\!-\!x\!-\!2\pi i\big)$

has infinitely many complex zeros, moreover I cannot get their exact values, but only an approximation of them.

I have also tried to put a parameter $b$ in the integral:

$I(b)=\displaystyle\int_{-\infty}^{+\infty}\frac{b}{\left(e^x-1-x\right)^2+b^2}~\mathrm dx$

and find a relation between $I(b)$ and its derivative $I’(b)$, but in this way I get

$I’(b)=\dfrac1bI(b)-\displaystyle\int_{-\infty}^{+\infty}\frac{2b^2}{\left[\left(e^x-1-x\right)^2+b^2\right]^2}~\mathrm dx$

and I do not know how I can solve it in order to obtain $I(b)$.

Another attempt I made is to write the integrand function as a series and then integrate it, but I did not even manage to do it.

By using numerical methods I got that the result is approximately $\,0.33333\,,$ but I would like to obtain the result exactly which should be $\dfrac13\,.$

Could anyone give me a hint to calculate the integral ?

  • $\begingroup$ I remember ever solved this integral before. $\endgroup$
    – MathFail
    Apr 30, 2023 at 22:27
  • $\begingroup$ @MathFail, could you tell me how you solved it ? $\endgroup$
    – Angelo
    Apr 30, 2023 at 22:35
  • $\begingroup$ Sure, let me write it down. $\endgroup$
    – MathFail
    Apr 30, 2023 at 22:40

1 Answer 1


$$\begin{align} I&=\displaystyle\int_{-\infty}^{+\infty}\frac{\mathrm dx}{\big(e^x-1-x\big)^2+4\pi^2}=\frac{1}{4\pi i}(I_1-I_2)\\ \\ &I_1=\displaystyle\int_{-\infty}^{+\infty}\frac{1}{\big(e^x-1-x\big)-2\pi i}dx\;,\\&I_2=\int_{-\infty}^{+\infty}\frac{1}{\big(e^x-1-x\big)+2\pi i}dx\;. \end{align}$$

For $I_1$, let $z=x+2\pi i$, and for $I_2$, let $z=x-2\pi i$

$$I_1=\lim_{R\rightarrow\infty}\displaystyle\int_{-R+2\pi i}^{R+2\pi i}\frac{1}{e^z-1-z}dz\;,\\I_2=\lim_{R\rightarrow\infty}\int_{-R-2\pi i}^{R-2\pi i}\frac{1}{e^z-1-z}dz\;.$$

Note that

$$\lim_{R\rightarrow\infty}\int_{-R-2\pi i}^{-R+2\pi i}\frac{1}{e^z\!-\!1\!-z}dz=\lim_{R\rightarrow\infty}\int_{R-2\pi i}^{R+2\pi i}\frac{1}{e^z\!-\!1-\!z}dz=0$$

We have:

$$I_1-I_2=-\oint_C \frac{1}{e^z-1-z}dz=-2\pi i \cdot\mathrm{Res}$$

where the contour $C$ is the infinite long box region with the vertices:

$(R, 2\pi i), (-R, 2\pi i), (-R, -2\pi i), (R, -2\pi i)$, oriented counter-clockwisely.

Inside the contour, $(0, 0)$ is the solely second order pole, so the residue is:

$$\mathrm{Res}=\lim_{z\rightarrow 0}\frac{d}{dz}\left(z^2\cdot\frac{1}{e^z-1-z}\right)=-\frac{2}{3}$$


$$I=\frac{1}{4\pi i}\cdot (-2\pi i)\cdot \left(-\frac{2}3\right)=\frac{1}3 $$

  • $\begingroup$ Thank you very much for your help. $\endgroup$
    – Angelo
    Apr 30, 2023 at 23:14
  • $\begingroup$ you are welcome $\endgroup$
    – MathFail
    Apr 30, 2023 at 23:34
  • $\begingroup$ Is there a way to calculate I without contour ? $\endgroup$
    – Fjaclot
    May 3, 2023 at 14:53
  • $\begingroup$ I belive it must exist, but I couldn't find currenlty :) $\endgroup$
    – MathFail
    May 3, 2023 at 17:59

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