= Solution
Integrate $F(z)=(1+z^2)^{-(k+1)}$ along the interval $[-R,R]$ closed by the positively oriented upper semicircle, with $R>1$. There is just one enclosed <pole>, at $i$, of order $k+1$. There,
$$
h(z)=(z-i)^{k+1}F(z)=(z+i)^{-(k+1)}.
$$
Differentiating $k$ times gives
$$
h^{(k)}(z)=(-1)^k\frac{(2k)!}{k!}(z+i)^{-(2k+1)}.
$$
By part (b),
$$
\operatorname{res}(F,i)=\frac{(-1)^k(2k)!}{(k!)^2(2i)^{2k+1}}=\frac{(2k)!}{2^{2k+1}i(k!)^2}.
$$
On the semicircle $|1+z^2|\geq R^2-1$, so its integral has absolute value at most $\pi R/(R^2-1)^{k+1}$, tending to zero. The real integral converges absolutely, so the <residue theorem> and the limit $R\to\infty$ give
$$
\boxed{\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^{k+1}}=2\pi i\operatorname{res}(F,i)=\pi\frac{(2k)!}{(k!)^2}4^{-k}}.
$$
Back to article page
