# I need some help with the integral $\int_0^\infty \frac{x^3}{e^x+1} \, dx$

I've been stuck on an integral for days now and would love to get some help with it:

$$\int_0^\infty \frac{x^3}{e^x+1} \, dx$$

My teacher was also "kind" enough to give me the answer to another similiar integral:

$$\int_0^\infty \frac{x^3}{e^x-1}dx=\frac{\pi^4}{15}$$

So that I should (based on that answer) calculate the integral:

$$\int_0^\infty \frac{x^3}{e^x+1} \, dx$$

I can't see how we can simplify/substitute anything so that it'll match the "help integral" so that we can use it's value to calculate the actual integral.

When $$A=\int\frac{x^3}{e^{x}+1}dx,\ B= \int\frac{x^3}{e^{x}-1}dx$$, then $$A-B = \int\frac{-2x^3\,dx}{e^{2x}-1} =\int\frac{- t^3/4}{e^t-1} \frac{dt}{2} =-\frac{1}{8}B$$

• How did you know to subtract the integrals? – gen-z ready to perish May 17 at 5:54
• Did you mean why I consider $A-B$ ? – HK Lee May 17 at 6:00
• precisely ${}{}{}{}{}$ – gen-z ready to perish May 17 at 6:33

Note that $$\dfrac{x^3}{e^x+1} - \dfrac{x^3}{e^x-1} = x^3 \cdot \dfrac{-2}{e^{2x}-1} = -\dfrac{1}{4} \dfrac{(2x)^3}{e^{2x}-1},$$ so $$\begin{gather}\int_0^\infty \dfrac{x^3}{e^x+1} \, dx - \int_0^\infty \dfrac{x^3}{e^x-1} \, dx = \\[6pt] =-\dfrac{1}{4} \int_0^\infty \dfrac{(2x)^3}{e^{2x}-1}\,dx = -\dfrac{1}{8} \int_0^\infty \dfrac{(2x)^3}{e^{2x}-1}\,d(2x) = -\dfrac{1}{8} \int_0^\infty \dfrac{u^3}{e^{u}-1}\,du.\end{gather}$$

An approach using series.

Note that $$f(x)=\frac{x^3}{e^x+1}=\frac{x^3}{2e^{x/2}}\text{sech }\frac{x}2$$ And then recall that for $$x>0$$ $$\text{sech }x=-2\sum_{k\geq1}(-1)^ke^{(1-2k)x}$$ So $$f(x)=\sum_{k\geq1}(-1)^{k+1}x^3e^{-kx}$$ Hence $$J=\int_0^\infty\frac{x^3}{e^x+1}dx=\sum_{k\geq1}(-1)^{k+1}\int_0^\infty x^3e^{-kx}dx$$ The final integral: \begin{align} q_k&=\int_0^\infty x^{3}e^{-kx}dx\\ &\overset{kx\mapsto x}=\frac1{k^4}\int_0^\infty x^{4-1}e^{-x}dx \end{align} Then recall the definition of the Gamma function: $$\Gamma(s)=(s-1)!=\int_0^{\infty}x^{s-1}e^{-x}dx\qquad \text{Re }s>0$$ Hence $$q_k=\frac6{k^4}$$ Which gives $$J=6\sum_{k\geq1}\frac{(-1)^{k+1}}{k^4}=6\eta(4)=\frac{7\pi^4}{120}$$ Where $$\eta(s)$$ is the Dirichlet Eta Function.

In fact, we can show that $$\int_0^\infty \frac{x^s}{e^x+1}dx=(1-2^{2-s})\Gamma(s+1)\zeta(s-1)$$ With the same approach.

If you're interested in an approach to get both integrals without knowing either, $$\int_0^\infty\frac{x^3e^{-x}dx}{1\pm e^{-x}}=\sum_{n\ge 1}(\mp 1)^{n-1}\int_0^\infty x^3e^{-nx}dx=6\sum_{n\ge 1}\frac{(\mp 1)^{n-1}}{n^4},$$giving either $$6\zeta(4)$$ or $$6\eta(4)$$.

Integration by parts with $$u=x^3$$ and $$dv=\frac {dx}{e^x+1}$$ may start you on the right track.

• Could you elaborate a little on your approach? With $dv=\frac {dx}{e^x+1}$ becoming $x-ln(e^x+1)$ multiplied with a $3x^2$ behind the integral sign when performing integration by parts, how is that becoming easier? – imranfat May 16 at 17:22