= Solution
For $x>0$ and $\alpha>1$, the preceding antiderivative gives
$$
\frac{\operatorname{sech}x-\operatorname{sech}(\alpha x)}x=\int_1^\alpha\operatorname{sech}(xy)\tanh(xy)\,dy.
$$
The integrand is nonnegative, so <Tonelli's theorem> permits reversing the two integrals without first assuming absolute integrability. Put $u=xy$ in the inner integral to obtain the <hyperbolic-secant Frullani integral>:
$$
I=\int_1^\alpha\frac{dy}{y}\int_0^\infty\operatorname{sech}u\tanh u\,du=\int_1^\alpha\frac{dy}{y}=\boxed{\ln\alpha}.
$$
Here $\int_0^\infty\operatorname{sech}u\tanh u\,du=[-\operatorname{sech}u]_0^\infty=1$. Near zero, cancellation makes the original integrand $O(x)$; at infinity it decays exponentially. The finite result therefore also establishes ordinary convergence of the <improper integral>.
Back to article page
