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Is the integral :

$ \frac{1}{{2}\mathit{\pi}{i}}\mathop{\int}\limits_{L}{\frac{{e}^{z}}{{z}^{2}\mathrm{{+}}{a}^{2}}} $

If contained the closed disk $ \left|{z}\right|\mathrm{\leq}{a} $

Is equal to :

$ \frac{1}{{2}\mathit{\pi}{i}}{\mathrm{(}}\frac{{e}^{\mathrm{{-}}{ai}}}{\mathrm{{-}}{2}{ai}}\mathrm{{+}}\frac{{e}^{ai}}{2ai}{\mathrm{)}} $

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    Or should be equal to zero – 2017-01-26

1 Answers 1

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Here is my guess. I don't think you can conclude that z0 is zero if z0 is simply on or inside the contour. That would be Cauchy-Goursat's theorem. The reason for this is because we are dealing with an integrand that is not analytic at the points where the denominator is zero.

So yes, we want to use Cauchy's Integral Formula. For this formula you have to look to see where the integrand is not analytic. This happens at z=+ai and z=-ai. The problem is that both of these points are actually on the disk |z|<=a.

What I think you have to do now is split this contour into a union of two contours (C1, C2) where each contour encloses or is on each non-analytic point separately. So z=+ai belongs to C1 and z=-ai belongs to C2.

Now you have to evaluate two integrals and add their results together. You will have to also algebraicly manipulate each integral's denominator. For example they should look like:

C1 contour integrand is e^z/(z-ai)/(z+ai) and C2 contour integrand is e^z/(z+ai)/(z-ai).

Now use Cauchy's Integral Formula since we now have a simple pole with degree 1. The formula is 2(pi)(i)f(z0).

My final result is:

e^(-ai)/(-2ai)+e^(ai)/(2ai)

So yes, your posted solution answer is correct.

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    If the OP really meant the Cauchy principal value, then you can [evaluate it by taking the mean](https://en.wikipedia.org/wiki/Cauchy_principal_value#Formulation) of the two contour integrals $\int_{|z| = a+\epsilon}f(z)dz,\int_{|z| = a-\epsilon}f(z)dz$ – 2017-01-26