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

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

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.