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Does this integral $(1)$

$$\int_{0}^{\infty}{e^x-1\over e^{ax}-1} dx=F(a)\tag1$$

$a\ge1$

has a closed form?

We may rewrite it as

$$\int_{0}^{\infty}{1\over e^{x(a-1)}-e^{-x}}-\int_{0}^{\infty}{1\over e^{ax}-1} dx=F(a)\tag2$$

we have $$\zeta(s)={1\over \Gamma(s)}\int_{0}^{\infty}{x^{s-1}\over e^{x}-1}dx\tag3$$

Guess of a closed form for $F(a)$

$$F(a)=\color{red}{{\pi\over 2a}\cot\left({\pi\over a}\right)}-{1\over a}\ln(2a)+ {2\over a}\sum_{n=1}^{\left\lfloor {a-1\over 2}\right\rfloor}\cos\left({2n\pi\over a}\right)\ln\sin\left({n\pi\over a}\right)\tag4$$

I was checking on the wolfram integral, the red part give part of the closed form but the black part seem to be a mistake somewhere I can't figured it out.

According to wolfram integral:

$$F(2)=\ln(2)$$

$$F(3)={1\over 2}\ln(3)-{\sqrt{3}\over 18}\pi$$

$$F(4)={3\over 4}\ln(2)-{\pi\over 8}$$

user91500
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gymbvghjkgkjkhgfkl
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  • $$\int_{0}^{\infty}{e^x-1\over e^{ax}-1} dx=\sum_{n=2}^\infty\frac{\zeta(n)}{a^n}$$ – Did Sep 19 '17 at 20:29
  • See https://math.stackexchange.com/questions/505573/infinite-series-sum-limits-n-2-infty-frac-zetankn – user91500 Sep 21 '17 at 12:11

1 Answers1

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\begin{align*} \int_{0}^{\infty} \frac{e^x-1}{e^{ax}-1} \, dx &= \frac{1}{a}\int_{0}^{\infty} \frac{e^{x/a}-1}{e^{x}-1} \, dx \\ &= \frac{1}{a} \sum_{n=1}^{\infty} \frac{1}{n!a^n} \int_{0}^{\infty} \frac{x^n}{e^{x}-1} \, dx \\ &= \frac{1}{a} \sum_{n=1}^{\infty} \frac{1}{n!a^n} \Gamma(n+1)\zeta(n+1) \\ &= \sum_{n=1}^{\infty} \frac{\zeta(n+1)}{a^{n+1}}, \end{align*}

which confirms @Did's answer. Also, $\text{(2)}$ is not valid as both integral diverges to infinity. An alternative expression can be found by applying the substitution $t = e^{-x}$:

\begin{align*} \int_{0}^{\infty} \frac{e^x-1}{e^{ax}-1} \, dx &= \frac{1}{a}\int_{0}^{\infty} \frac{e^{x/a}-1}{e^{x}-1} \, dx \\ &= \frac{1}{a}\int_{0}^{1} \frac{t^{-1/a}-1}{1-t} \, dt \\ &= - \frac{1}{a}\left(\gamma + \psi\left(1 - \frac{1}{a}\right) \right), \end{align*}

where $\psi$ is the digamma function.

Sangchul Lee
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