Complex integration with pole singularity at zero of order three

Question

Find the integral .

$$\frac{1}{2\pi i}\int_{\gamma} \frac{1}{z^2(e^{z}-1)}dz$$ where $\gamma=\{z: z=\frac{1}{2}e^{2\pi i t},0\leq t\leq1 \}$

Here $z=0$ is a pole singularity of order 3. Using Cauchy Integral formula leads to complicated calculation. How to proceed?

Answer 1

Let$$f(z)=\begin{cases}\frac z{e^z-1}&\text{ if }z\neq0\\1&\text{ if }z=0.\end{cases}$$Then $f$ is analytic and your integral is equal to$$\frac1{2\pi i}\int_\gamma\frac{f(z)}{z^3}\,\mathrm dz.\tag1$$Now, if $f(z)=a_0+a_1z+a_2z^2+\cdots$, then$$a_2=\frac{f''(0)}2=(1).$$On the other hand,\begin{align}z&=(e^z-1)(a_0+a_1z+a_2z^2+\cdots)\\&=\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)(a_0+a_1z+a_2z^2+\cdots).\end{align}So,$$\left\{\begin{array}{l}a_0=1\\\frac{a_0}2+a_1=0\\\frac{a_0}6+\frac{a_1}2+a_2=0\end{array}\right.$$and this will give you $a_2$ and therefore the value of $(1)$.

Answer 2

Use the formula $g''(0)=\frac {2!} {2\pi i} \int_{\gamma} \frac {g(z)} {z^{3}}dz$ where $g(z)=\frac z {e^{z}-1}$ if $z \neq 0$ and $g(0)=1$. The formula is valid because $g$ is analytic.