Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 17

Use integration by substitution with $u=\ln x$, so $du=dx/x$. For $x>1$, also $u>0$, andThe domain condition explains why no absolute value is needed around the inner logarithm.

Separate one factor of the hyperbolic sine and use the hyperbolic Pythagorean identity ${\sinh }^{2}x={\cosh }^{2}x-1$. With $u=\cosh x$ and $du=\sinh xdx$, integration by substitution givesFor real $x$, the hyperbolic cosine never vanishes. Differentiating the result gives $\sinh x(1-{\mathrm{sech}}^{2}x)={\sinh }^{3}x/{\cosh }^{2}x$, checking the sign of the reciprocal term.

Use integration by parts in $I_{n+1}$, writing the integrand as $x^{2n+1}(x{e}^{-x^2})$. Since an antiderivative of $x{e}^{-x^2}$ is $-e^{-x^2}/2$,The boundary term vanishes because the Gaussian exponential function decays faster than any fixed power. These improper integrals are absolutely convergent, justifying the limiting integration by parts.

Starting from the permitted Gaussian integral $I_0=\sqrt{\pi }$, the Gaussian moment recurrence gives $I_1=\sqrt{\pi }/2$ and thenMore generally, iteration gives $I_n=(2n-1)!!\sqrt{\pi }/2^n$, using the double factorial with $(-1)!!=1$.