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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.

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  • $\begingroup$ 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? $\endgroup$
    – DonAntonio
    Commented Apr 5, 2013 at 3:25
  • $\begingroup$ @DonAntonio Yeah, I know about the winding number stuff. $\endgroup$
    – Alexander Gruber
    Commented Apr 5, 2013 at 3:27

1 Answer 1

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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 $$I_{\alpha}(z) = \sum_{m=0}^{\infty} \dfrac1{m! \Gamma(m+\alpha+1)} \left(\dfrac{z}2 \right)^{2m+\alpha}$$

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