It is well-known that: $$ \int_{0}^{+\infty}\frac{x}{e^{x}-1}\,dx = \zeta(2) = \sum_{n\geq 1}\frac{1}{n^2} \tag{1}$$

but what is known about $$ I_2 = \int_{0}^{+\infty}\frac{x^2}{e^x-1-x}\,dx \tag{2} $$ or $$ I_n = \int_{0}^{+\infty}\frac{x^n}{e^x-\sum\limits_{k=0}^{n-1}\frac{x^k}{k!}}\,dx \tag{3}$$ ?

They seem pretty hard to tackle through the usual geometric series approach or the residue theorem, but it would be interesting to find some sharp bounds, too. An approximation that came to my mind is, for instance: $$ I_n \approx \int_{0}^{+\infty} n! e^{-\frac{z}{n+1}}\,dz = (n+1)!$$ but it does not seem so tight. A better approximation is: $$ \exp\left(-\frac{x}{n+1}-\frac{x^2}{2(n+1)^2}\right)\leq \frac{\frac{x^n}{n!}}{\sum_{m>n}\frac{x^m}{m!}}\leq\exp\left(-\frac{x}{n+1}-\frac{nx^2} {2(n+1)^2(n+2)}\right)$$ but that just improves the previous approximation by a constant factor.

The previous inequality has a nice side effect. From $\frac{x}{e^x-1}\approx\exp\left(-\frac{x}{2}-\frac{x^2}{24}\right)$ it follows that: $$ \int_{0}^{+\infty}\left(-\frac{1}{2}-\frac{x}{12}\right)\frac{x}{e^x-1}\,dx \approx \int_{0}^{+\infty}\left(-\frac{1}{2}-\frac{x}{12}\right)\exp\left(-\frac{x}{2}-\frac{x^2}{24}\right)\,dx = -1,$$ or:

$$ \frac{\zeta(2)}{2!}+\frac{\zeta(3)}{3!}\approx 1.$$

  • $\begingroup$ yes sorry and in the same way you should get $\displaystyle\int_0^\infty \frac{x^{s-1}}{e^x-x-1} dx = \int_0^\infty \frac{x^{s-1} e^{-x}}{1-(x+1)e^{-x}} dx = \int_0^\infty x^{s-1} \sum_{k=1}^\infty e^{-kx}\sum_{m=0}^{k-1} {k-1 \choose m} x^m dx = \sum_{k=1}^\infty \sum_{m=0}^{k-1} k^{-s-m} {k-1 \choose m}\Gamma(s+m) = \ldots$ $\endgroup$ – reuns May 2 '16 at 11:11
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    $\begingroup$ maybe you can also use the residue theorem with a well-chosen contour, for example with $\displaystyle \int_0^\infty \frac{x^{s-1}}{e^x-x-1} dx = \int_{-\infty}^\infty \frac{e^{su}}{e^{e^u}-e^u -1} du$ and exploiting that $\displaystyle\frac{e^{s(u+2\pi)}}{e^{e^(u+2\pi)}-e^{(u+2\pi)} -1} = e^{2 \pi s} \frac{e^{su}}{e^{e^u}-e^u -1}$ $\endgroup$ – reuns May 2 '16 at 11:13
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    $\begingroup$ It is possible to prove that the first integral is equal to $$\sum_{k\geq0}\frac{e^{k+1}\Gamma(k+3,k+1)}{(k+2)(k+1)^{k+3}}-1.$$ So maybe we can find a good approximation from it. $\endgroup$ – Marco Cantarini May 18 '16 at 19:11
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    $\begingroup$ This might be of some help? You can see that your integral is obtained if $s=2$. $\endgroup$ – Eleven-Eleven May 20 '16 at 15:34
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    $\begingroup$ My pleasure. I do graduate research with one of the co-authors, Dr. Hassen, with methods of generalizations. I just happened to be looking at this paper yesterday. After you helped me on a previous post today I came across this post. Hope it works out for you. $\endgroup$ – Eleven-Eleven May 20 '16 at 15:43

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