Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 20

Assume real square-integrable functions on the same integration domain. Put $A=\int f^2$, $B=\int g^2$, and $C=\int fg$. If $B=0$, then $g=0$ almost everywhere, so $C=0$ and the claim follows. If $B>0$, nonnegativity of a square givesMultiplying by $B$ proves the integral form of the Cauchy-Schwarz inequality:For nonzero functions, equality holds exactly when they are proportional almost everywhere. The zero-function cases also give equality.

The integrand and both limits depend on $x$, so use the Leibniz integral rule. The upper and lower endpoint contributions and the interior derivative giveFor $x\ne 0$, an antiderivative in $t$ is $t\sin (xt)/x+\cos (xt)/x^2$. Evaluating it and collecting terms givesAt $x=0$ the original Leibniz expression has zero endpoint terms and an integral over a zero-length interval, giving $\boxed{F^{\prime }(0)=0}$. The displayed quotient has a removable singularity and the same limiting value.

The parameter derivative of the integrand is $\cos (a\log x)$, whose absolute value is bounded by one on $0<x<1$. This integrable bound justifies differentiation under the integral sign, givingThe original integral also exists: $|\sin (a\log x)/\log x|\le |a|$, and its apparent singularity at $x=1$ is removable. Set $x=e^{-t}$ to evaluate the derivative:ConsequentlySince $I(0)=0$, integration in the parameter fixes the constant and gives the logarithmic sine integral identity