Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2009 / Analysis / 5 / Solution

For $t>0$, integrate $e^{itz}/(1+z^2)$ over the upper semicircle of radius $R$. The arc tends to zero, since $|e^{itz}|\le 1$ there and the denominator is of order $R^2$. The sole pole is $i$, of residue $e^{-t}/(2i)$, so the real part gives $I(t)=\pi e^{-t}$. The integrand defining $I$ is even in $t$, and $I(0)=\pi$ by the arctangent integral. Hence

For $J$ use $e^{itz}/[z(1+z^2)]$ with a small clockwise semicircular indentation above zero. Its residue at zero is one, so the small arc tends to $-i\pi$. The pole $i$ has residue $-e^{-t}/2$, and the large arc again vanishes. ThusThe imaginary part is the ordinary convergent integral $J(t)$: its apparent singularity at zero is removable, with limit $t$, and its tails are absolutely integrable. Therefore $J(t)=\pi (1-e^{-t})$ for $t>0$. Oddness in $t$ and the value zero at $t=0$ give

Differentiation is valid at every real $t$, including zero. Indeed the mean value theorem bounds the absolute difference quotient of $\sin (tx)/[x(1+x^2)]$ with respect to $t$ by $1/(1+x^2)$, an integrable majorant independent of $t$. Dominated convergence therefore gives $\boxed{J^{\prime }(t)=I(t)\text{ for all }t\in \mathbb{R}}$. In particular both one-sided derivatives at zero equal $\pi$.