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I want to compute the following integral $$\oint_{|z|=1}\frac{\exp \left (\frac{1}{z} \right)}{z^2-1}\,dz$$ The integrand has essential singularity at the origin, and $2$-poles at $\pm 1$,which lie on the curve $|z|=1$ so I can't apply residue formula. How can I proceed?

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the integral doesn't exits in the usual sense; if you want the principal value (which is most probable) then you should count the two residues on the circle with $1/2$ (it works for simple poles – user8268 Nov 28 '12 at 20:44
since this is a part of a bigger exercise with parameters ( here a=1 in the denominator) and the text says that for some values of $a$ the integral doesn't exists, i think this is the case. But how can i justify this? simply saying that residue formula doesn't apply? – bateman Nov 28 '12 at 21:12

1 Answer 1

The integral as posed does not exist. No Cauchy principal value will help this. I can, however, pose a problem that is very close that will produce a pretty nifty result.


$$\oint_C dz \frac{e^{1/z}}{z^2-1} $$

where $C$ is the following contour:

enter image description here

i.e., the circle $|z|=1$ with semicircular notches of radius $\epsilon$ cut into the circle at the poles $z=\pm 1$. By Cauchy's theorem, this integral is zero. However, we can use this fact to deduce a nontrivial integral.

The contour integral is also equal to

$$i \int_{-\pi/2}^{-\epsilon} d\theta \, e^{i \theta} \frac{e^{e^{-i \theta}}}{e^{i 2 \theta}-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{1/(1+\epsilon e^{i \phi})}}{(1+\epsilon e^{i \phi})^2-1} \\ + i \int_{\epsilon}^{\pi-\epsilon} d\theta \, e^{i \theta} \frac{e^{e^{-i \theta}}}{e^{i 2 \theta}-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{1/(-1+\epsilon e^{i \phi})}}{(-1+\epsilon e^{i \phi})^2-1} \\ + i \int_{\pi+\epsilon}^{3 \pi/2} d\theta \, e^{i \theta} \frac{e^{e^{-i \theta}}}{e^{i 2 \theta}-1} $$

Note that the point at which I started and ended the integration on the circle is arbitrary; I chose it so that I was able to encompass the small semicircles in their entirety rather than split them up. Now take the limit as $\epsilon \to 0$, and I can simply express the Cauchy principal value over the circle over any interval of length $2 \pi$. By Cauchy's theorem, we have

$$i PV \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{e^{e^{-i \theta}}}{e^{i 2 \theta}-1} - i \frac{\pi}{2} e + i \frac{\pi}{2} e^{-1} = 0 $$

Now separate out real and imaginary parts. With a little manipulation, we may conclude the following:

$$\int_0^{2 \pi} d\theta \frac{e^{\cos{\theta}} \sin{(\sin{\theta})}}{\sin{\theta}} = \pi \left ( e-\frac1{e}\right ) $$

$$PV \int_0^{2 \pi} d\theta \frac{e^{\cos{\theta}} \cos{(\sin{\theta})}}{\sin{\theta}} = 0$$

Sometimes it is interesting where these contour integrals can take us!


It is not trivial that the contour integral above is zero. To show this, we need to expand the integrand in a Laurent series about $z=0$:

$$\frac{e^{1/z}}{z^2-1} = \left ( 1+\frac1{z} + \frac12 \frac1{z^2} + \cdots \right ) \frac1{z^2} \left (1+\frac1{z^2} + \cdots \right ) = \frac1{z^2} + \cdots$$

As there are no terms in $1/z$, the residue of this function at $z=0$ is zero, as is the integral.

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