Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2004 / 5

Integrate $F(z)=e^{iz}/(z^2+1)$ around the real segment from $-R$ to $R$ and the counterclockwise upper semicircle, with $R>1$. The only enclosed pole is $i$, whose residue is $e^{-1}/(2i)$. On the semicircle, $|e^{iz}|\le 1$ and $|z^2+1|\ge R^2-1$, so its integral has magnitude at most $\pi R/(R^2-1)\to 0$.

The residue theorem thus gives ${\int }_{-\infty }^{\infty }{e}^{ix}/(x^2+1)dx=\pi e^{-1}$. The line integral converges absolutely, so taking its real part is justified:

For a nonnegative integer $n$, put $z=e^{i\theta }$ on the unit circle. Since $\cos \theta =(z+z^{-1})/2$ and $d\theta =dz/(iz)$,The residue at zero is the coefficient of $z^{2n}$ in $(1+z^2{)}^{2n}$, namely $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. The residue theorem givesThe contour integrand has no other poles. The formula also includes $n=0$, giving $2\pi$.