# Evaluate $\int_{0}^{\infty} \frac{1}{\sqrt{x(1+e^x)}}\mathrm dx$

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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Ok, maybe the exercise consisted only in proving the integral exists, which is easy. Thanks! – Chon Sep 25 '11 at 12:47
A complex analysis approach might also be doable: the integrand has poles at $x_n = (2n + 1)\pi i$ – Gerben Sep 25 '11 at 13:14

\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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This is the same as Sasha's answer, just rewriting $\binom{-1/2}{n}$. – robjohn Sep 25 '11 at 13:30

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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