Studying for my complex-analysis exam I found an interesting integral. The activity consisted of calculating a complex integral around a curve using the residue theorem. The integral had the form $$ \oint_{\gamma}e^{z\over z-a} \cdot \mathbb{d}z $$ where $z, a \in \mathbb{C}$, and $a$ is an interior point of the closed curve $\gamma$.
I didn't know how to solve it using pure residue theorem because the function doesn't have the correct form to use it, so I decided to do it this way:
First, using the MacLaurin series for $e^x$ and the integral properties, I wrote: $$ \oint_{\gamma}\sum_{n=0}^\infty{\frac{z^n\over (z-a)^n}{n!}} \cdot \mathbb{d}z = \sum_{n=0}^\infty{\frac{1}{n!} \cdot \oint_{\gamma}{z^n\over (z-a)^n}} \cdot \mathbb{d}z$$
For $n=0$, because of the Cauchy integral theorem, the integral takes the form $\oint_{\gamma}{{z^0\over (z-a)^0}\cdot \mathbb{d}z} = \oint_{\gamma}{\mathbb{d}z} = 0$, so the first term of the sum will be $0$, which means that we can start the sum in $n=1$ without modifying the result.
Then, using the residue theorem for each integral term I got:
$$ \sum_{n=1}^\infty{\frac{1}{n!} \cdot \frac{2\pi i}{(n-1)!}\cdot \lim_{z\rightarrow a}{\left[ \frac{\mathbb{d}^{(n-1)} }{\mathbb{d} z^{(n-1)}}(z^n) \right]}} $$
Since $\frac{\mathbb{d}^{(n-1)} }{\mathbb{d} z^{(n-1)}}(z^n) = n!\cdot z$
$$ \sum_{n=1}^\infty{\frac{1}{n!} \cdot \frac{2\pi i}{(n-1)!}\cdot \lim_{z\rightarrow a}{\left(n!\cdot z \right)}} = \sum_{n=1}^\infty{ \frac{2\pi i}{(n-1)!}\cdot a} $$
So we got the following expression: $$ 2\pi i a\cdot\sum_{n=1}^\infty{ \frac{1}{(n-1)!}} $$
Now, replacing $k = n-1$, so that when $n=1$, $k=0$:
$$ 2\pi i a\cdot\sum_{k=0}^\infty{ \frac{1}{k!}} $$
where $\sum_{k=0}^\infty{ \frac{1}{k!}}$ is the MacLaurin series for $e$. Then, the initial integral around a curve $\gamma$ would be: $$ \oint_{\gamma}e^{z\over z-a} \cdot \mathbb{d}z = 2\pi i a e $$
I think it is an interesting result and I wanted to share it in the forum. However, I don't have the real result of the integral so I'm not actually sure that my method is correct, and I didn't see any example of an integral with this form in any book neither.
So my questions are:
Is my method correct for calculating this type of integrals?
Is there a better way of calculating this integral using the residue theorem?
Not sure what you mean by the "pure" residue theorem. Let's use $$\int_\gamma f(z)\,dz=2\pi i(\hbox{sum of residues inside $\gamma$})\ ,$$ where the residue at $z_0$ is by definition the coefficient of $(z-z_0)^{-1}$ in the Laurent series of $f(z)$ which converges for $0<|z-z_0|<r$. In this case the function is holomorphic everywhere except at $a$ and the relevant series is $$e^{z/(z-a)}=e^{1+a/(z-a)}=\sum_{n=0}^\infty \frac{e}{n!}\Bigl(\frac{a}{z-a}\Bigr)^n\ , \qquad 0<|z-a|<\infty\ .$$ The coefficient of $(z-a)^{-1}$ is $ea$ and the integral is $2\pi iea$.