= Hyperbolic-secant Frullani integral
{title2=$\int_0^\infty[\operatorname{sech}x-\operatorname{sech}(\alpha x)]\,dx/x=\ln\alpha$}
For $\alpha>1$, write $[\operatorname{sech}x-\operatorname{sech}(\alpha x)]/x=\int_1^\alpha\operatorname{sech}(xy)\tanh(xy)dy$. The integrand is nonnegative, so <Tonelli's theorem> permits reversal. Substitution $u=xy$ makes the inner $x$ integral $1/y$ because $\int_0^\infty\operatorname{sech}u\tanh u\,du=1$. 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.
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