Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 13

Use integration by substitution with $x=2u$ and the Gaussian being an even function:The factor two from the differential exactly cancels the half-line factor in the given full-line Gaussian integral.

Retain the original PDF's full-line integration limits. For $n\ge 2$, write $x^n{e}^{-x^2}dx=-x^{n-1}d(e^{-x^2})/2$. Integration by parts givesGaussian decay dominates every power, so the endpoint term vanishes, soThis is the full-line Gaussian moment recurrence; its initial values differ from the half-line version.

The integrand for $I_1$ is an odd function with absolute convergence, so $I_1=0$. For $I_2$, the recurrence gives $I_2=I_0/2$, where the supplied Gaussian integral is $I_0=\sqrt{\pi }$. HenceThe zero first moment is specific to the symmetric full-line domain.

Use integration by substitution with the reciprocal variable $t=1/x$, so $dx=-dt/t^2$ and $\log x=-\log t$. The endpoints $x=1,\infty$ become $t=1,0$. ThusRenaming the dummy integration variable proves the identity without evaluating either side.

The integral converges absolutely: near zero it behaves like $\log x$, and at infinity like $(\log x)/x^2$. Let its value be $J$. The integration by substitution $x=1/t$ givesThereforeAbsolute convergence makes this cancellation between the two ends legitimate.

The tangent half-angle substitution $t=\tan (x/2)$ gives $\sin x=2t/(1+t^2)$ and $dx=2dt/(1+t^2)$. The endpoints become zero and one, soUse completing the square, $t^2+t+1=(t+1/2{)}^2+3/4$, and integrate:Thus both integrals equal $\boxed{\pi /(3\sqrt{3})}$.