Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 20

Under the regularity and integrability conditions for the Leibniz integral rule,The first two terms account for the moving endpoints; the last accounts for the integrand's own parameter dependence.

For $m>-1$, integrate the base power to obtain $J(m)={\int }_0^1{x}^{m}dx=1/(m+1)$. Since ${\partial }_m^n{x}^m=x^m(\log x{)}^n$, differentiating $n$ times givesHere $n$ is a nonnegative integer. For justification, keep $m$ in a small closed interval bounded below by $m_0>-1$. The differentiated integrands are dominated by $x^{m_0}|\log x{|}^n$, whose integral becomes ${\int }_0^{\infty }{e}^{-(m_0+1)t}{t}^{n}dt$ under $x=e^{-t}$ and is finite. The dominated convergence theorem establishes the needed differentiation under the integral sign and the logarithmic moments of a power on the unit interval.

Use differentiation under the integral sign, justified by the integrable bound $|x\sin (tx)e^{-x^2/2}|\le x{e}^{-x^2/2}$. This givesIntegration by parts makes the boundary term vanish: sine vanishes at zero and the Gaussian tends to zero at infinity. ThereforeThe integrating factor $e^{t^2/2}$ makes $(e^{t^2/2}F{)}^{\prime }=0$, so $F(t)=C{e}^{-t^2/2}$. At $t=0$, the given half-line Gaussian integral fixes $C=\sqrt{\pi /2}$. ThusThis is the even cosine-transform form of the Fourier transform of a Gaussian.