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I am trying to evaluate

$$I = \int_{0}^{\infty} \frac{e^{-x} - 1 + x - \frac{x^2}{2}}{x^2 (e^x - 1)} \, dx$$

The integral has terms which suggest connections to harmonic series expansions, which are often associated with the Euler-Mascheroni Constant:

$$\gamma = \int_{0}^{\infty} \left(\frac{1}{e^x - 1} - \frac{1}{x}\right)e^{-x} \, dx$$

I tried substitutions like $x = t^n$ and comparisons with known integrals involving $ln(x)$ and zeta functions, without any success.

Are there any known techniques, substitutions, or expansions that might reveal its dependence on $\gamma$? Could this integral be computed directly using special functions (e.g., polylogarithms, zeta functions)?

mattTheMathLearner
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    Slight corrections ) Using analytical continuation and choosing $\lambda>1$ $$I(\lambda)=\int_0^\infty \frac{e^{-x} - 1 + x - \frac{x^2}{2}}{x^2 (e^x - 1)}x^\lambda , dx=\int_0^\infty\left(-x^{\lambda-2}e^{-x}+\frac{x^{\lambda-1}}{e^x-1}-\frac12\frac{x^\lambda}{e^x-1}\right)dx$$ $$=-\Gamma(\lambda-1)+\zeta(\lambda)\Gamma(\lambda)-\frac12\zeta(1+\lambda)\Gamma(1+\lambda)=-\frac{\Gamma(1+\lambda)}{\lambda(\lambda-1)}+\frac{\zeta(\lambda)\Gamma(1+\lambda)}\lambda-\frac12\zeta(1+\lambda)\Gamma(1+\lambda)$$ – Svyatoslav Jan 01 '25 at 14:16
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    Using $\Gamma(1+\lambda)=1-\gamma\lambda+O(\lambda^2),,, \zeta(\lambda)=-\frac12-\frac{\ln(2\pi)}2\lambda+O(\lambda^2),,,\zeta(1+\lambda)=\frac1\lambda+\gamma+o(1)$ and leading $\lambda\to 0$ $$I=I(\lambda=0)=1-\frac\gamma2-\frac{\ln(2\pi)}2=-0.207546...$$ – Svyatoslav Jan 01 '25 at 14:16
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    Thank you @Svyatoslav, this is wonderful – mattTheMathLearner Jan 01 '25 at 14:17

1 Answers1

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Recalling the Taylor series for $e^x$ we get \begin{align*} \int_{0}^{\infty} \frac{e^{-x} - 1 + x - \frac{x^2}{2}}{x^2 (e^x - 1)} \, \mathrm{d}x & = \int_{0}^{\infty} \frac{\sum_{k \color{blue}{\ge 0}} \frac{(-1)^k x^k}{k!} -\sum_{k= \color{blue}{0}}^{\color{blue}{2}} \frac{(-1)^k x^k}{k!}}{x^2 (e^x -1) }\, \mathrm{d}x\\ & = \sum_{k \color{blue}{\ge 3}}\frac{(-1)^k}{k!}\int_{0}^{\infty} \frac{x^{k-2}}{e^x-1}\, \mathrm{d}x\\ & \overset{\color{purple}{k-1 \to k}}{=} -\sum_{k \ge 2}\frac{(-1)^k}{k(k+1)} \zeta(k) \end{align*} where we use that $\int_{0}^{\infty} \frac{x^n}{e^x -1}\mathrm{d}x =\Gamma(n+1)\zeta(n+1)$ for $n >0$ on the final step. Lastly, knowing that $\sum_{k \ge 2}\frac{(-1)^k}{k(k+1)} \zeta(k) = \frac12\left(\ln(2) + \ln(\pi)+ \gamma -2 \right)$ we get the closed form given by @Svytoslav in the comments.

Robert Lee
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