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When I was watching Vtuber content, I found this guy has an integral formula in his header of twitter.

$$\int \cfrac{150u^3}{e^{\pi u}-1}du$$

It seemed there is no attempt to Evaluate this integral on the Internet, so I wanted to try it myself.

My Attempt:

Let $x = \pi u$ , so $\cfrac{dx}{du}= \pi$ and $du=\cfrac{dx}{\pi}$.

also, $u=\cfrac{x}{\pi}$.

Rewritten the integral by $x$ and $dx$, We can write:

$$\int \cfrac{150u^3}{e^{\pi u}-1}du = \int \cfrac{150(\cfrac{x}{\pi})^3}{e^{x}-1}\times \cfrac{dx}{\pi}= \int \cfrac{150\cfrac{x^3}{\pi^3}}{\pi\times(e^{x}-1)}dx=\int \cfrac{150x^3}{\pi^4(e^{x}-1)}dx$$ And I stucked here. I also thinking about partial integration, but I have no clue.

How can I Evaluate this from here?

Infinity_hunter
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R.A
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  • I tried $e^{x}=t$ or $e^{x}-1=t$ but I couldn't replace $150\times x^{3}$ – R.A Aug 20 '21 at 05:34
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    without the useless factor of 150: $$\frac{12 \pi u \left(\pi u \text{Li}_2\left(e^{\pi u}\right)-2 \text{Li}_3\left(e^{\pi u}\right)\right)+24 \text{Li}_4\left(e^{\pi u}\right)+\pi ^3 u^3 \left(4 \log \left(1-e^{\pi u}\right)-\pi u\right)}{4 \pi ^4}$$. where $Li$ is the polyLog function. – David G. Stork Aug 20 '21 at 05:40
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    As shown in the above comment, the integral is most likely a joke. – Parcly Taxel Aug 20 '21 at 05:42
  • As is said, you should never take an integral and try it unless you're sure it is solveable. – ultralegend5385 Aug 20 '21 at 05:50

1 Answers1

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I suspect there is no closed-form in elementary functions for this indefinite integral because of a good reason.

The Rieman Zeta function and Gamma function are related to each other by this integral $$\zeta(s) \Gamma(s) = \int_0^{\infty}\frac{x^{s-1}}{e^x -1 } \, dx$$ where $\Re(s) > 1$

So we get $$\int_0^{\infty} \frac{x^3}{e^x -1} \, dx = \zeta(4)\Gamma(4) = \frac {\pi^{15}}{15}$$

However, we can have a power series representation using $$\frac{x^3}{e^x -1} = \sum_{m = 0}^{\infty} \frac{B_mx^{m+2}}{m!}$$ where $B_m$ are Bernouli's number.

Infinity_hunter
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