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I'm wondering if this integral can be expressed in some compact form:

$$ \int\limits_{0}^{\infty} x^{\frac{3}{2}}\frac{1}{1 + e^x}dx $$

And if not - why?

I was thinking that it was somehow connected to the Riemann $ \zeta $ function, because:

$$ \zeta(s) = \frac{1}{\Gamma(s)}\int\limits_0^{\infty} \frac{x^{s-1}}{e^x - 1} \text{dx} $$

but I'm not sure how to transform it to obtain such form.

I would appreciate a hint

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  • $\begingroup$ $\eta(s) = (1-2^{1-s}) \zeta(s)$ and your integral is $\eta(5/2)$. by the way, all these formulas are based on $n^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-nx} dx$ $\endgroup$ – reuns Feb 1 '16 at 20:40
  • $\begingroup$ @user1952009 modulo a multiplicative constant ($3\sqrt{\pi}/4$ or so). $\endgroup$ – mickep Feb 1 '16 at 20:45
  • $\begingroup$ @mickep his integral is $\eta(5/2) \Gamma(5/2)$ and you're right $\Gamma(5/2) = 3/4 \sqrt{\pi}$ $\endgroup$ – reuns Feb 1 '16 at 20:51
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Collect $e^x$ and arranging a bit to obtain:

$$\int_0^{+\infty} x^{3/2}\frac{1}{e^x(e^{-x} + 1)}\ \text{d}x$$

Now the term $\frac{1}{1 + e^{-x}}$ ca be written as a geometric series, since $e^{-x} < 1$, so then:

$$ \int_0^{+\infty} x^{3/2} e^{-x}\sum_{k = 0}^{+\infty}\ (-e^{-x})^k\ \text{d}x\ =\ \sum_{k = 0}^{+\infty}\ (-1)^k \int_0^{+\infty} x^{3/2}e^{-x(1+k)}\ \text{d}x $$

For the moment you may call $a = 1+k$ in order to have a less messy function, and then try to integrate that function.

This is a hint and despite the integration is not immediate, it's not difficult either. Keep in mind that you'll get some Special function in return so I hope you're good in calculus!

Try it, I'll post the continue in 1 day, if you need it.

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