Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S1 / 3

For $t>0$, close the real-axis contour in the upper half-plane: $|e^{izt}|=e^{-t\mathrm{Im}z}$ supplies decay. The only enclosed simple pole is $z=i$, with residue $e^{-t}/(2i)$. Thus $h(t)=2\pi i{e}^{-t}/(2i)=\pi e^{-t}$.

For $t<0$, close downwards. The clockwise orientation contributes $-2\pi i$ times the residue $e^t/(-2i)$ at $z=-i$, givingAt $t=0$, direct integration of $1/(1+p^2)$ gives $\pi$, consistent with the continuous extension. The large arc tends to zero by Jordan's lemma for nonzero $t$, or the denominator estimate when $t=0$.

Let $I$ denote the requested integral. Symmetry and the period of ${\sin }^2\theta$ make the corresponding integral over $0\le \theta \le 2\pi$ equal to $4I$. With $z=e^{i\theta }$, this becomesThe denominator factors as $-2(z^2-1/2)(z^2-2)$. The two poles inside the unit circle are $z=\pm 1/\sqrt{2}$. Each residue is $1/(6i)$, because the derivative of the polynomial denominator there is $6z$. The residue theorem gives $4I=2\pi /3$, so