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$$\int_{-\infty}^{\infty}\frac{1}{\left(e^{t}-t\right)^{2}+\pi^{2}}dt=\frac{1}{1+\Omega}$$

I have recently become interested in the W-Lambert function which has led me to ask about this identity. I know the integral is evaluated on this post but I am curious about only real-valued methods.

I attempted to split the integral and then u-sub $u=e^t-t$ to get something related to arctan but things got really weird with the Lambert W function and I was unable to make any progress.

Dylan Levine
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1 Answers1

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Here is an approach that avoids contour integration, although it is strongly motivated by complex-analytic approaches.

Let $I$ denote the integral. We claim that $I - \frac{1}{1+\Omega}$ is zero. Indeed, define $f(z)$ by

$$ f(z) = \frac{1}{\Omega + 1} \cdot \frac{1}{z + \Omega} - \frac{1}{e^z + z}. $$

Then it is not hard to check that

  1. $f(z)$ has no singularity in the region $\mathcal{D} = \mathbb{R} \times [-\pi, \pi]$, and

  2. $f(z) \to 0$ as $\Re(z) \to \pm\infty$ in $\mathcal{D}$, and

  3. $ \iint_{\mathcal{D}} |f'(x + iy)| \, \mathrm{d}x\mathrm{d}y < \infty $.

Graph of f(z)

This function is relevant to the integral $I$ by the following relation:

$$ \frac{1}{(e^x - x)^2 + \pi^2} - \frac{1}{(\Omega + 1)[(x + \Omega)^2 + \pi^2]} = \frac{f(x + i\pi) - f(x - i\pi)}{2\pi i} $$

Integrating both sides with respect to $x$, we get

$$ \begin{align*} I - \frac{1}{\Omega + 1} &= \int_{-\infty}^{\infty} \left[ \frac{1}{(e^x - x)^2 + \pi^2} - \frac{1}{(\Omega + 1)[(x + \Omega)^2 + \pi^2]} \right] \, \mathrm{d}x \\ &= \frac{1}{2\pi i} \int_{-\infty}^{\infty} [f(x + i\pi) - f(x - i\pi)] \, \mathrm{d}x \\ &= \frac{1}{2\pi i}\int_{-\infty}^{\infty} \int_{-\pi}^{\pi} i f'(x + iy) \, \mathrm{d}y\mathrm{d}x \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \int_{-\infty}^{\infty} f'(x + iy) \, \mathrm{d}x\mathrm{d}y \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \bigl[ f(x + iy) \bigr]_{x=-\infty}^{x=\infty} \, \mathrm{d}y \\ &= 0. \end{align*} $$

Sangchul Lee
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  • nice answer! but does using "i" count as a real-valued method? – Maxime Jaccon Mar 01 '25 at 20:59
  • @MaximeJaccon, Since there is no complex integration involved, we can take real part to every quantity and every integral appearing in my answer to turn it into a purely real-analytic solution. This, however, will obfuscate the key idea of the solution, so I did not choose to do so. – Sangchul Lee Mar 01 '25 at 23:05
  • Is there a way to evaluate the integral directly with a similar method? – Тyma Gaidash Mar 01 '25 at 23:17
  • @ТymaGaidash, I can tweak my answer to give a solution that does not require knowing the value $\frac{1}{\Omega+1}$ in advance. However, it still relies on complex-analytic way of thinking (with real-analysis disguise). I'd also love to see a solution that purely relies on real-analysis. – Sangchul Lee Mar 02 '25 at 05:06