 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/8/c/ii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

For

x > 0

and

α > 1

, the preceding antiderivative gives

\frac{sech x - sech ( αx )}{x} = \int_{1}^{α} sech (x y) tanh (x y) d y .

The integrand is nonnegative, so Tonelli's theorem permits reversing the two integrals without first assuming absolute integrability. Put

u = x y

I = \int_{1}^{α} \frac{d y}{y} \int_{0}^{\infty} sech u tanh u d u = \int_{1}^{α} \frac{d y}{y} = ln α .

Here

\int_{0}^{\infty} sech u tanh u d u = [- 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.

 New to topics? Read the docs here!