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My Problem

I'm trying to solve problems from old qualifying exams in complex analysis and right now I'm stuck on the following exercise.

Compute the complex integral

$ \int_{\gamma} e^{\frac{1}{z^2 - 1}}\sin{\pi z} \,\mathrm dz $

where $\gamma$ is a closed curve in the right half plane that has index $N$ with respect to the point $1$.


My attempt

By the Residue theorem, since the function $f(z) := e^{\frac{1}{z^2 - 1}}\sin{\pi z}$ has only an isolated singularity at $z = 1$ in the right half plane, which in this case is an essential singularity, we can compute the integral by finding the residue at $1$, more precisely

$ \int_{\gamma} e^{\frac{1}{z^2 - 1}}\sin{\pi z} \, \mathrm dz = 2\pi i N \operatorname{Res}{(f(z); 1)} $

where the $N$ comes from the assumption on the index of the curve.

Now my problem is that I can't compute this residue. The "obvious" things that I have tried are finding the Laurent expansions of the functions $e^{\frac{1}{z^2 - 1}}$ and $\sin{\pi z}$ around $z = 1$. It is easy to find that

$ \sin{\pi z} = -\frac{\pi}{1!}(z - 1) + \frac{\pi^3}{3!}(z - 1)^3 -\frac{\pi^5}{5!}(z - 1)^5 + \frac{\pi^7}{7!}(z - 1)^7 + \cdots $

Then I tried doing the following with the exponential:

$ e^{\frac{1}{z^2 - 1}} = \sum_{n = 0}^{\infty}\frac{1}{n!}\frac{1}{(z^2 - 1)^n} $

and I thought that maybe then expressing the fraction $\frac{1}{z^2 - 1}$ as

$ \frac{1}{z^2 - 1} = \frac{1}{2} \left ( \frac{1}{z-1} - \frac{1}{z+1} \right ) $

and using the Laurent expansion

$ \frac{1}{z+ 1} = \frac{1}{z - 1 + 2} = \frac{1}{2}\frac{1}{1 + \frac{z - 1}{2}} = \frac{1}{2} \sum_{n = 0}^{\infty}\frac{(z - 1)^n}{2^n} $

could be of help.

But now I don't see much hope of this working because I would have to put this last infinite series back into the series for the exponential and there's an $n$-th power there, and finally to top it all I would have to multiply by the Laurent series for the $\sin{\pi z}$ to try to get a hold of the coefficient of $\frac{1}{z - 1}$.


I would really appreciate any help with this. Thanks.

  • 0
    just out of interest, what is the solution to this problem (if known)?2016-09-09

1 Answers 1

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Using your series expansions we can write: $ e^{\frac1{z^2-1}} \sin \pi z = \sum_{n,k \ge 0} (-1)^{k+1} \frac{\pi^{2k+1}}{(2k+1)!\,n!} \frac1{(z+1)^n}\frac1{(z-1)^{n-2k-1}} $ Now (for any $z$ inside the circle of radius 2 centered at $z=1$) $ (z+1)^{-n} = \sum_{l\ge0} \frac{(-n)(-n-1)\cdot\ldots\cdot(-n-l+1)}{l!}2^{-n-l} (z-1)^l $ So in total we get $ e^{\frac1{z^2-1}} \sin \pi z = \sum_{n,k,l \ge 0} (-1)^{k+1} \frac{\pi^{2k+1}}{(2k+1)!\,n!} {-n \choose l} \frac1{2^{n+l}} \frac1{(z-1)^{n-2k-l-1}} $ The coefficient of $\frac1{z-1}$ is then $ \sum_{n,k \ge 0} \frac{(-1)^{k+1} \pi^{2k+1}}{(2k+1)!\,n!\,4^{n-k}} {-n \choose n-2k} $ There must be a nicer solution however....