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?
Answer
$$ \begin{align} \int_0^\infty\frac{1}{\sqrt{x(1+e^x)}}\mathrm{d}x &=2\int_0^\infty\frac{1}{\sqrt{1+e^{x^2}}}\mathrm{d}x\\ &=2\int_0^\infty(1+e^{-x^2})^{-1/2}e^{-x^2/2}\;\mathrm{d}x\\ &=2\int_0^\infty\sum_{k=0}^\infty(-\tfrac{1}{4})^k\binom{2k}{k}e^{(2k+1)x^2/2}\;\mathrm{d}x\\ &=\sum_{k=0}^\infty(-\tfrac{1}{4})^k\binom{2k}{k}\sqrt{\frac{2\pi}{2k+1}} \end{align} $$
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