# Limit of a complex valued integral

The question is:

Compute $$\lim_{p\rightarrow0^{+}}\int_{C_p}\frac{e^{3iz}}{z^{2}-1}dz$$

Where $$C_p: z = 1 + pe^{i\theta}$$

My initial thought was to use residues, yet the poles are -1 and 1, so they're on the real line (thus the Residue Theorem does not apply). My next thought was to find some way to make the integral work with the Cauchy Integral Formula, but I can't find a way to do that since a partial fraction decomposition won't work in this case. So, I am stuck.

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Note that the path is a circle of radius $p$ centered on the real $1$ so that the residues may be used. – Raymond Manzoni Jul 25 '12 at 23:19

Your contour $C_p$ is a circle of radius $p \to 0$ centred at the point $z=1$. This means that there is a singularity in the contour (not on its path). Because of this, we may use residue theorem (at singularity $z=1$) to evaluate this integral. If

$$f(z)=\frac{\exp(3iz)}{z^2-1}=\frac{\exp(3iz)}{(z+1)(z-1)}$$

we see that

$$\operatorname{Res}_{z=1}f(z)=\lim_{z \to 1} (z-1) \frac{\exp(3iz)}{(z+1)(z-1)} = \frac{\exp(3i)}{2}$$

Then, by residue theorem

$$\oint_{C_p} f(z)\, dz=2\pi i \operatorname{Res}_{z=1} (f(z)) = 2 \pi i \frac{\exp(3i)}{2}=\pi i \exp(3 i)$$

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Thanks guys. For some reason I was getting hung up on the typical display of a half circle centered at 0 for residues (what they generally show as examples in the Residue Theorem of complex analysis books). Guess I should've thought outside the box and made a different Jordan curve that didn't run over the singularities!!! – Payton Jul 25 '12 at 23:39
It's not "thinking outside the box": they told you what the contour was. – Robert Israel Jul 26 '12 at 0:35