# Evaluating the (complex) integral $\int_\gamma \frac{e^{z+z^{-1}}}{z}\mathrm dz$ using residues.

I am trying to evaluate the following integral.

$$\int_\gamma \frac{e^{z+z^{-1}}}{z}\mathrm dz$$ where $\gamma$ is the path $\cos(t)+2i\sin(t)$ for $0\leq t <4\pi$.

So, $\gamma$ is an ellipse running twice counterclockwise around $0$, which is where the function has a singularity. I'm sure I need to use the residue theorem to evaluate this.

1. (for homework) I'm not good with the Residue theorem yet. Can I get a road map for the canonical solution to this problem? (i.e. the way I'm "probably supposed to" do it.) I can work through the details myself.

2. (non-homework) Is it possible to solve this problem with the Laurent series approach from this answer using the residues for $e^z/z$ and $e^{-z}$? (or $e^z/\sqrt{z}$ and $e^{-z}/\sqrt{z}$, if that would be better.)

To be clear about where I'm confused for part (1): I see that the hypothesis for the Residue theorem is met: the above function is analytic with an isolated singularity at $0$, we're goin around it twice, so $\int_\gamma f=4\pi i \operatorname{Res}(0,f)$. But from here I don't know how to perform the computations.

• I'm afraid you're going to need not only the residue theorem but also in fact its "heavy" version, with the winding number. Have you already studied this? Commented Apr 5, 2013 at 3:25
• @DonAntonio Yeah, I know about the winding number stuff. Commented Apr 5, 2013 at 3:27

The only singularity for the integrand is at $z=0$, which is within the contour of integration. The integral is nothing but $$2 \pi i \cdot \left(\text{Residue at } z=0 \text{ of }\left(\dfrac{e^{z+1/z}}z \right) \right) \cdot \text{Number of times the closed curve goes about the origin}$$ Let us write the Laurent series about $z=0$. We then get $$e^{z+1/z} = e^z \cdot e^{1/z} = \sum_{k=0}^{\infty} \dfrac{z^k}{k!} \cdot \sum_{m=0}^{\infty} \dfrac1{z^m \cdot m!}$$ Hence, $$\dfrac{e^{z+1/z}}z = \dfrac{e^z \cdot e^{1/z}}z = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \dfrac{z^{k-m-1}}{k! m!}$$ The term $z^{-1}$ in the series is when $k=m$. Hence, the coefficient of $\dfrac1z$ is $$\sum_{k=0}^{\infty} \dfrac1{(k!)^2}$$ Hence, your answer is $$4 \pi i \sum_{k=0}^{\infty} \dfrac1{(k!)^2} = 4 \pi iI_0(2)$$where $I_{\alpha}(z)$ is the modified Bessel's' function of the first kind given by $$I_{\alpha}(z) = \sum_{m=0}^{\infty} \dfrac1{m! \Gamma(m+\alpha+1)} \left(\dfrac{z}2 \right)^{2m+\alpha}$$