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I would like to evaluate: $$ \int_{0}^{\infty} \frac{1}{\sqrt{x(1+e^x)}}\mathrm dx $$ Given that I can't find $ \int \frac{1}{\sqrt{x(1+e^x)}}\mathrm dx $, a substitution is needed: I tried $$ u=\sqrt{x(1+e^x)} $$ and $$ u=\frac{1}{\sqrt{x(1+e^x)}} $$ but I could not get rid of $x$ in the new integral.... Do you have ideas of substitution?

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    $\begingroup$ Ok, maybe the exercise consisted only in proving the integral exists, which is easy. Thanks! $\endgroup$ – Chon Sep 25 '11 at 12:47
  • $\begingroup$ A complex analysis approach might also be doable: the integrand has poles at $x_n = (2n + 1)\pi i$ $\endgroup$ – Gerben Sep 25 '11 at 13:14
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$$ \begin{align} \int_0^\infty\frac{1}{\sqrt{x(1+e^x)}}\mathrm{d}x &=2\int_0^\infty\frac{1}{\sqrt{1+e^{x^2}}}\mathrm{d}x\\ &=2\int_0^\infty(1+e^{-x^2})^{-1/2}e^{-x^2/2}\;\mathrm{d}x\\ &=2\int_0^\infty\sum_{k=0}^\infty(-\tfrac{1}{4})^k\binom{2k}{k}e^{(2k+1)x^2/2}\;\mathrm{d}x\\ &=\sum_{k=0}^\infty(-\tfrac{1}{4})^k\binom{2k}{k}\sqrt{\frac{2\pi}{2k+1}} \end{align} $$

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    $\begingroup$ This is the same as Sasha's answer, just rewriting $\binom{-1/2}{n}$. $\endgroup$ – robjohn Sep 25 '11 at 13:30
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I doubt this integral admits closed-form expression for its value. You could expand integrand into a series:

$$ \frac{1}{\sqrt{1+\mathrm{e}^x}} = \frac{\mathrm{e}^{-x/2}}{\sqrt{1+\mathrm{e}^{-x}}} = \sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} \mathrm{e}^{-\left(n+\frac{1}{2}\right)x} $$

You can then integrate term-wise to get that

$$ \int_0^\infty \frac{1}{\sqrt{x \left(1+\mathrm{e}^x \right)}} \mathrm{d} x = \sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} \sqrt{ \frac{2 \pi }{ 2n+1}} \simeq 2.0343 $$

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