How to solve $\int_{0}^{\infty}\frac{x^3 e^{-x}}{1-{e^{-x}}} dx$?

Question

Is there any obscure technique that should be used? When integrating by parts, I got a "matryoshka" leading nowhere.

Also, how does WolframAlpha approach it? It gives a precise solution: $\frac{\pi^4}{15}$, but no steps available.

Answer 1

we know that: $$\sum_{n=1}^{\infty} e^{-nx} = \frac{e^{-x}}{1 - e^{-x}} $$ so: $$ \int_{0}^{\infty} \frac{x^3 e^{-x}}{1 - e^{-x}} \, dx =\sum_{n=1}^{\infty} \int_{0}^{\infty} x^3 e^{-nx} \, dx$$ We also know that: $$\int_{0}^{\infty} x^a e^{-nx} \, dx = n^{-1-a} \Gamma(a+1) \quad \text{for} \ \Re(a) > -1 \ \text{and} \ \Re(n) > 0 $$ so: $$\sum_{n=1}^{\infty} \int_{0}^{\infty} x^3 e^{-nx} \, dx = \sum_{n=1}^{\infty} \frac{6}{n^4}$$ $$= 6\zeta(4) $$ $$= \frac{\pi^4}{15} $$

Answer 2

Consider the geometric series

$$\sum_{n=1}^{\infty} e^{-x n}=\frac{e^{-x}}{1-e^{-x}}$$

Try using the above formula as a starting point, can you complete the rest?

Answer 3

Just another solution.

$$I=\int\frac{x^3 e^{-x}}{1-{e^{-x}}}\, dx$$ $$e^{-x}=t \quad \implies \quad x=-\log(t)\quad \implies \quad dx=-\frac{dt} t$$ $$I=\int \frac{\log ^3(t)}{1-t}\,dt$$ A few integrations by parts give $$I=-\log(1-t) \log ^3(t)-3 \text{Li}_2(t) \log ^2(t)+6 \text{Li}_3(t) \log (t)-6 \text{Li}_4(t)$$

$$J=\int_0^\infty \frac{x^3 e^{-x}}{1-{e^{-x}}}\, dx=-\int_0^1 \frac{\log ^3(t)}{1-t}\,dt=\frac{\pi ^4}{15}$$

Answer 4

We can generalize this integral: $$I(s)=\int_0^\infty \frac{x^{s-1}} {e^x-1}\mathrm{d}x=\int_0^\infty \frac{x^{s-1}e^{-x}} {1-e^{-x}}\mathrm{d}x$$ and prove that it is equal to $\zeta(s)\Gamma(s)$. We start by expanding using a geometric series: $$\frac{1}{1-e^{-x} }=\sum_{k=0}^{\infty}e^{-kx}$$ Therefore, $$I(s) = \int_0^\infty x^{s-1}e^{-x}\sum_{k=0}^{\infty}e^{-kx}\mathrm{d}x=\sum_{k=1}^{\infty}\int_0^\infty x^{s-1}e^{-kx}\mathrm{d}x$$ Next, we use the substitution $t=kx$ : $$I(s) =\sum_{k=1}^{\infty}\int_0^\infty \left(\frac{t}{k}\right)^{s-1}e^{-t}\,\frac{\mathrm{d}t}{k}=\sum_{k=1}^{\infty}\frac{1}{k^s}\int_0^\infty t^{s-1}e^{-t}\mathrm{d}t$$ Observe that the integral in the sum is independent of $k$ and equal to $\Gamma(s)$:$$I(s) =\Gamma(s)\sum_{k=1}^{\infty}\frac{1}{k^s}=\Gamma(s)\zeta(s)$$ Now, we can set $s=4$ to find the value of the original integral: $$\int_0^\infty \frac{x^{3}e^{-x}} {1-e^{-x}}\mathrm{d}x=I(4)=\Gamma(4)\zeta(4)=\frac{\pi^4}{15}$$