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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.

First post on this page, so please be kind! :)

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

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    $\begingroup$ How did you know to subtract the integrals? $\endgroup$ – let's have a breakdown May 17 at 5:54
  • $\begingroup$ Did you mean why I consider $A-B$ ? $\endgroup$ – HK Lee May 17 at 6:00
  • $\begingroup$ precisely ${}{}{}{}{}$ $\endgroup$ – let's have a breakdown May 17 at 6:33
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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 $$\begin{gather}\int_0^\infty \dfrac{x^3}{e^x+1} \, dx - \int_0^\infty \dfrac{x^3}{e^x-1} \, dx = \\[6pt] =-\dfrac{1}{4} \int_0^\infty \dfrac{(2x)^3}{e^{2x}-1}\,dx = -\dfrac{1}{8} \int_0^\infty \dfrac{(2x)^3}{e^{2x}-1}\,d(2x) = -\dfrac{1}{8} \int_0^\infty \dfrac{u^3}{e^{u}-1}\,du.\end{gather} $$

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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.

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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 $6\eta(4)$.

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

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  • $\begingroup$ 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? $\endgroup$ – imranfat May 16 at 17:22

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