(1) Prove that$$\int_0^\infty \frac{(\mathrm{e}^{-ax}-\mathrm{e}^{-bx})^2}{x^2}\mathrm{d}x=\ln\frac{(2a)^{2a}(2b)^{2b}}{(a+b)^{2(a+b)}}$$
(2) Prove that$$\int_0^\infty \frac{\mathrm{e}^{-ax^2}-\mathrm{e}^{-bx^2}}{x^2}\mathrm{d}x=\frac{1}{2}\ln\frac{b}{a} $$
I just learned improper integral with parameters and worked out a basic example below
$$
\int_0^\infty \frac{\mathrm{e}^{-ax}-\mathrm{e}^{-bx}}{x}\mathrm{d}x=\ln\frac{b}{a}
$$
But I do not know how to compute the two integrals above. I think the basic example may be of some help and tried integration by parts, inspired by $\int_0^\infty\frac{\sin x}{x}\mathrm{d}x$ and $\int_0^\infty\frac{\sin^2 x}{x^2}\mathrm{d}x$.
Any hint is appreciated.
Best Answer
Write $\frac{1}{x^2}=\int_0^\infty ye^{-xy}dy$ so the LHS of (1) is$$\begin{align}&\int_0^\infty dy\:y\int_0^\infty dx(e^{-(2a+y)x}-2e^{-(a+b+y)x}+e^{-(2b+y)x})\\&=\int_0^\infty dy\:y\left(\frac{1}{2a+y}-\frac{2}{a+b+y}+\frac{1}{2b+y}\right).\end{align}$$You can handle (2) the same way, thought it helps to first replace $\int_0^\infty$ with $\tfrac12\int_{-\infty}^\infty$, so you just have to evaluate Gaussians on $\Bbb R$.