Complex Methods | Part IB, 2002

A function f(z)f(z) has an isolated singularity at aa, with Laurent expansion

f(z)=n=cn(za)nf(z)=\sum_{n=-\infty}^{\infty} c_{n}(z-a)^{n}

(a) Define res (f,a)(f, a), the residue of ff at the point aa.

(b) Prove that if aa is a pole of order k+1k+1, then

res(f,a)=limzah(k)(z)k!, where h(z)=(za)k+1f(z).\operatorname{res}(f, a)=\lim _{z \rightarrow a} \frac{h^{(k)}(z)}{k !}, \quad \text { where } \quad h(z)=(z-a)^{k+1} f(z) .

(c) Using the residue theorem and the formula above show that

dx(1+x2)k+1=π(2k)!(k!)24k,k1\int_{-\infty}^{\infty} \frac{d x}{\left(1+x^{2}\right)^{k+1}}=\pi \frac{(2 k) !}{(k !)^{2}} 4^{-k}, \quad k \geq 1

Typos? Please submit corrections to this page on GitHub.