= 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 gives
$$
\frac{\partial I}{\partial\alpha}=-\int_0^\infty xe^{-\alpha x^2}\,dx=-\frac1{2\alpha},\qquad\frac{\partial I}{\partial\beta}=\int_0^\infty xe^{-\beta x^2}\,dx=\frac1{2\beta}.
$$
Thus $I=-\tfrac12\ln\alpha+\tfrac12\ln\beta+C$, with $C$ independent of both parameters. Since $I(\alpha,\alpha)=0$, $C=0$, and
$$
\boxed{I(\alpha,\beta)=\frac12\ln\frac\beta\alpha.}
$$
The 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 $1/2$.
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