Is $\int_{|z|=2}\exp\bigl(e^{1/z}\bigr)\,dz = 2\pi ie$

Question

Question is to find $$\int_{|z|=2}\exp\bigl(e^{1/z}\bigr)\,dz$$

Cauchy residue theorem says

If $f(z)$ is regular, except at a finite number of poles within a closed contour $C$ and continuous on the boundary of $C$, then $ \int_{C} f(z)\,dz=2\pi i \sum R$, where $\sum R$ is the sum of the residues of $f(z)$ at its poles within $C$.

In the above integral $f(z)= \exp\bigl(e^{1/z}\bigr)$ has essential singularity at $z=0$.

Also residue for a function at a point $a$ is the coefficient of $\frac{1}{z}$ in its Laurent series expansion about $a$.

If we write the Laurent series expansion for $\exp\bigl(e^{1/z}\bigr)$, then the coefficient of $\frac{1}{z}$ is $e$ and therefore by Cauchy's residue theorem the value of the integral is $2\pi ie$.

But is this correct? In residue theorem we have residue at poles but here $z=0$ is an essential singularity.

Help me find my mistake. Thank you

Answer 1

The residue theorem is not restricted to functions with poles. A precise statement can be found in Wikipedia: Residue Theorem:

Let $U$ be a simply connected open subset of the complex plane containing a finite list of points $a_1, \ldots, a_n$, and $f$ a function defined and holomorphic on $U \setminus \{a_1, ..., a_n\}$. Let $\gamma $ be a closed rectifiable curve in $U$ which does not meet any of the $a_k$. Then $$ \int_\gamma f(z) \, dz = 2 \pi i \sum_{k=1}^n \operatorname{I}(\gamma, a_k) \operatorname{Res}(f,a_k) $$ where $\operatorname{I}(\gamma, a_k)$ denotes the winding number of $\gamma $ around $a_k$.

$f$ has “isolated singularities” at each $a_k$, that can be poles, essential singularities, or removable singularities.

If $\gamma $ is a simple positive-oriented closed curve then the formula simplifies to $$ \int_\gamma f(z) \, dz = 2 \pi i \sum_{k} \operatorname{Res}(f,a_k) $$ where the sum is now taken over all $a_k$ which are “inside” $\gamma$.

In your case $f(z) = \exp\bigl(e^{1/z}\bigr)$ is holomorphic in $\Bbb C \setminus \{ 0 \}$, and therefore $$ \int_{|z|=2}\exp\bigl(e^{1/z}\bigr)\,dz = 2 \pi i \operatorname{Res}(f, 0) = 2 \pi i e \, , $$ i.e. your result is correct.