 Solution

First check the improper integral exists: near zero its numerator is $(\beta -\alpha )x^2+O(x^4)$, so the quotient is $O(x)$; at infinity both exponentials decay. For positive parameters in compact neighborhoods, the differentiated integrands have integrable bounds, justifying differentiation under the integral sign. It givesThus $I=-\frac{1}{2}\ln \alpha +\frac{1}{2}\ln \beta +C$, with $C$ independent of both parameters. Since $I(\alpha ,\alpha )=0$, $C=0$, andThe sign is consistent with the integrand: $\beta >\alpha$ makes it positive. Neither separate integral ${\int }_0^{\infty }{e}^{-\alpha x^2}dx/x$ converges at zero; it is the difference that is meaningful. The substitution $u=x^2$ identifies the result as a Frullani integral with an extra factor $1/2$.