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For $0<a<b$, the nonnegative function $e^{-xy}{1}_{\{x>0,a\le y\le b\}}$ is measurable and has integral ${\int }_a^{b}dy/y=\log (b/a)$. Tonelli theorem first proves integrability without assuming it; Fubini theorem then permits the order of integration to be exchanged. Integrating in $y$ gives the stated Frullani integral. The cancellation between the exponentials is essential at $x=0$.

The Fubini theorem for an integrable Borel-measurable function states that, if ${\int }_{{\mathbb{R}}^2}|f|<\infty$, its sections are integrable for almost every coordinate andFor a nonnegative measurable function, Tonelli theorem gives the same equalities with infinity allowed before integrability is established.

Here $f$ is the continuous exponential multiplied by the indicator of a Borel set, so it is Borel-measurable and nonnegative. Tonelli and integration in $x$ giveThis proves integrability. We may now integrate in the reverse order; for $x>0$ the inner integral is $(e^{-ax}-e^{-bx})/x$. Hence the Frullani integral isThe apparent singularity at zero is removable in the integrand's limiting value $b-a$.

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 $1/2$.