While reading a paper, I encountered the following result.
$$\int_0^\infty \frac{1}{\sqrt{2\pi \mathcal{E}x}} \log_e \left(1+\frac{1}{\xi^2 x}\right)dx = \sqrt{\frac{2\pi}{\xi^2\mathcal{E}}}$$
The integrand diverges near $0$ so I believe it is to be understood as a limit from above there.
However when I attempted to prove it (by substitution and by parts), I couldn't succeed. If this result is at all true, perhaps it may follow from contour integration (The $2\pi$ in RHS is somewhat a tipoff). However I'm not sure how to approach the problem from that angle.
Can someone help me on this? If a solution is possible using elementary methods, please give me a hint on that.
The paper is "Capacity Results of an Optical Intensity Channel With Input-Dependent Gaussian Noise" by Stefan Moser. Page 6 has this result.
It can be solved by elementary methods. First, make the substitution $y = \xi \sqrt{x}$; this factors the parameters out of the integral and leaves you with the integral $$ \int_0^\infty \ln \left( 1 + \frac{1}{y^2} \right) \, dy. $$ This latter integral can be solved via integration by parts—the indefinite integral has a closed form in terms of elementary functions (though applying these bounds of integration to the indefinite integral requires a bit of tricky limit-taking). Let me know if you need a further hint to see how to do this.