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On Mathworld one finds without proof the integral

$$\int_{-\infty}^{\infty} \frac{dx}{(\exp(x) - x)^2 + \pi^2} = \frac{1}{1 + W(1)}$$

where $W$ denotes the Lambert W function. How can one show this? The link given on Mathworld is broken.

user111187
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    If you refer to Moll's article, you may have a look at this. Unfortunately, it has no proof. – Jean-Claude Arbaut Aug 03 '14 at 20:04
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    The answer is here: http://math.stackexchange.com/questions/45745/interesting-integral-related-to-the-omega-constant-lambert-w-function – Lord Soth Aug 03 '14 at 20:15
  • @LordSoth Oh, thank you. – user111187 Aug 03 '14 at 20:26
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    @user111187 With the same approach, you may generalize this result to $$I(\alpha)= \displaystyle \int_{-\infty}^{+\infty}\displaystyle \frac{1}{(e^x-x+\alpha+\ln \alpha)^2+\pi^2}= \displaystyle \frac{1}{\alpha+1}, \quad \alpha > 0. $$ – Olivier Oloa Aug 03 '14 at 21:32

1 Answers1

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It is just an application of the residue theorem. Take $\gamma_R$ as the union of the path that goes straight from $-R$ to $R$ and the semicircle that goes from $R$ to $-R$, counter-clockwise oriented. We have: $$ \int_{-\infty}^{+\infty}\frac{dx}{(\exp(x)-x)^2+\pi^2}=\lim_{R\to +\infty}\int_{\gamma_R}f(z)\,dz=2\pi i\cdot\sum_{\xi\in S_R}\operatorname{Res}(f(z),z=\xi)$$ where $f(z)=\frac{1}{(\exp(z)-z)^2+\pi^2}$ and $S_R$ is the set of the zeroes of $(\exp(z)-z)^2+\pi^2$ having positive imaginary part and modulus bounded by $R$. For any $R$ big enough, $S_R$ is made of one element only, namely the solution of: $$ \exp(z)-z = -i\pi,$$ that is: $$\xi = i\pi - W(1).$$ The residue theorem hence gives: $$ \int_{-\infty}^{+\infty}\frac{dx}{(\exp(x)-x)^2+\pi^2} = \frac{1}{1+W(1)}$$ as wanted.

Jack D'Aurizio
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  • Why would $S_R$ be made of one element only for $R$ big enough? Aren't there infinitely many zeroes of $(e^x-x)^2+\pi^2$ in the upper half-plane, arbitrarily far away from $0$? – Franklin Pezzuti Dyer Dec 19 '17 at 20:30
  • @Nilknarf: indeed this old answer of mine is inaccurate, but there is a cancellation of residues which allows to prove that the only residue which really matters is the residue at $i\pi-W(1)$, as claimed here. For a fixed proof, have a look at robjohn's solution to the linked question. – Jack D'Aurizio Dec 19 '17 at 20:36
  • Okay, thank you and sorry to bother you! – Franklin Pezzuti Dyer Dec 19 '17 at 20:37