- Let \(F\) be an analytic function inside and on a simple closed curve \(C\), except for a pole of order \(m\geq 1\) at \(z=a\) inside \(C\). Prove that
\[\frac{1}{2 \pi i}\oint_{C} F(\tau) d\tau = \lim_{\tau\rightarrow a} \frac{d^{m-1}}{d\tau^{m-1}}\big((\tau-a)^m F(\tau))\big).\]
- Evaluate \[\oint_{C}\frac{e^{\tau}}{(\tau^2+\pi^2)^2}d\tau\] where \(C\) is the circle \(|z|=4\).
Solution
Part (a) is missing the standard factor \(1/(m-1)!\) and is false as printed when \(m>2\). The correct formula is \[ \frac{1}{2\pi i}\oint_C F(\tau)\,d\tau =\operatorname{Res}_{\tau=a}F =\frac1{(m-1)!}\lim_{\tau\to a} \frac{d^{m-1}}{d\tau^{m-1}} \bigl((\tau-a)^mF(\tau)\bigr). \] Indeed, if \[ F(\tau)=\sum_{k=-m}^{\infty}c_k(\tau-a)^k, \] then \((\tau-a)^mF(\tau)\) is holomorphic at \(a\), and its \((m-1)\)-st derivative at \(a\) is \((m-1)!c_{-1}\). The residue theorem gives \((2\pi i)^{-1}\oint_C F=c_{-1}\).
For part (b), the double poles inside \(|z|=4\) are \(z=\pm i\pi\). At \(i\pi\), \[ \operatorname{Res}_{z=i\pi} \frac{e^z}{(z^2+\pi^2)^2} =\left.\frac{d}{dz}\frac{e^z}{(z+i\pi)^2}\right|_{z=i\pi} =\frac1{4\pi^2}+\frac{i}{4\pi^3}. \] Similarly, \[ \operatorname{Res}_{z=-i\pi} \frac{e^z}{(z^2+\pi^2)^2} =\frac1{4\pi^2}-\frac{i}{4\pi^3}. \] Thus the sum of the residues is \(1/(2\pi^2)\), and \[ \boxed{\displaystyle \oint_C\frac{e^\tau}{(\tau^2+\pi^2)^2}\,d\tau =2\pi i\frac1{2\pi^2}=\frac{i}{\pi}.} \]