Hyperbolic-secant Frullani integral (${\int }_0^{\infty }[\mathrm{sech}x-\mathrm{sech}(\alpha x)]dx/x=\ln \alpha$)

For $\alpha >1$, write $[\mathrm{sech}x-\mathrm{sech}(\alpha x)]/x={\int }_1^{\alpha }\mathrm{sech}(xy)\tanh (xy)dy$. The integrand is nonnegative, so Tonelli's theorem permits reversal. Substitution $u=xy$ makes the inner $x$ integral $1/y$ because ${\int }_0^{\infty }\mathrm{sech}u\tanh udu=1$. Integrating $dy/y$ proves the formula. The original difference is $O(x^2)$ at zero, making division by $x$ integrable. For $0<\alpha <1$, substitution and reversal of the two terms give the same $\ln \alpha$ result; at $\alpha =1$ both sides vanish.