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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.

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+1 for such a neatly posed question. –  Pedro Tamaroff Aug 4 '12 at 3:02

2 Answers 2

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....

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Notice that what you are really interested in is the behavior of your function $e^{1/(z^2-1)}\sin(\pi z)$ around $z=1$. In particular, $\sin(\pi z)$ is nice and analytic everywhere, so there isn't much sense in expanding it in a Laurant series.It's the $e^{1/(z^2-1)}$ part that's carrying the (essential) singularity. More specifically, by your expansion, you can write $e^{\frac{1}{2(z-1)}}[e^{\frac{1}{2(z+1)}}\sin(\pi z)]$, and the brackets indicate the analytic part. Now, $e^{1/{2(z-1)}}=1+\frac{1}{2(z-1)}+\ldots$ which multiplies the bracketed analytic part. Then by Cauchy Residue theorem, you're going to take derivatives of $e^{\frac{1}{2(z+1)}}\sin(\pi z)$ at $z=1$. By product rule these look to work out nicely because the $\sin$ part is 0 except when it turns into a cosine from differentiation.

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First, $\sin(\pi z)/(z-1)\to-\pi$ as $z\to1$. Second, even accounting for the first, if we look at $$ \color{#C00000}{\left((z-1)e^\frac1{2(z-1)}\right)} \color{#00A000}{\left(\frac{\sin(\pi z)}{(z-1)}e^\frac1{2(z+1)}\right)}=\color{#C00000}{\left((z-1)+ \frac12+\frac1{8(z-1)}+\dots\right)} \color{#00A000}{\left(-\pi e^{1/4}+\dots\right)} $$ we are ignoring many terms combining terms like $\frac1{(z-1)^{k+1}}$ from the red series and terms like $(z-1)^k$ from the green series. –  robjohn Aug 4 '12 at 11:34
Echoing @robjohn, you need to look at the convolution of the respective Laurent series. –  copper.hat Aug 4 '12 at 19:57
@robjohn: thanks for the correction! I've edited but, I'm now not quite seeing why the analytic part needs a Laurant expansion? –  Alex R. Aug 4 '12 at 21:51
@Sam: you are looking for the coefficient of the $\frac1{z-1}$ term of the Laurent expansion of $e^{1/(z^2-1)}\sin(\pi z)$ at $z=1$. You get a contribution from $\frac1{8(z-1)}$ times $-\pi e^{1/4}$; however, you also get a contribution from the $\frac1{(z-1)^2}$ term in red times the $(z-1)$ term in green, and from the $\frac1{(z-1)^3}$ term in red times the $(z-1)^2$ term in green, and so on. –  robjohn Aug 5 '12 at 0:16
@MorganSherman: I'm not sure what you are saying. There are infinitely many negative powers of $z-1$ in red and infinitely many positive powers of $z-1$ in green. To compute the coefficient of $\frac1{z-1}$ in the Laurent series of the product, you have to account for each of the infinitely many products that yields $\frac1{z-1}$. –  robjohn Aug 5 '12 at 0:20

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