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March 12th, 2015, 09:39 AM   #1
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Application of Cauchy generalized integral theorem

I am asked to find $\displaystyle\oint_\gamma \frac{z^2 + e^z} {(z-i\pi)^2}dz$. Using Cauchy's Generalized Integral Formula.

I am getting $\boxed{ - 1+2\pi i}$.

Is this correct? (I don't have an answer key)

Also I need to find $\displaystyle\int_0^{2\pi} e^{\alpha\cos \theta} \sin(\alpha\cos \theta)d\theta$. Using Cauchy's integral formula. I am not quite sure how to do this one. My attempt was to apply Euler's formula and then go from there. How do I use Cauchy's integral formula?

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March 19th, 2015, 05:27 AM   #2
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1. Assuming $\displaystyle z=\pi i$ lies within the closed contour of the first problem, by the Cauchy integral formula, $\displaystyle \oint_{\gamma}\frac{z^2+e^z}{(z-\pi i)^2}=2\pi i f'(\pi i)$ where $\displaystyle f(\alpha)=\alpha^2+e^\alpha$. Therefore, the answer is,

$\displaystyle 2\pi i f'(\pi i)=2\pi i(2(\pi i)+e^{\pi i})=-4\pi^2-2\pi i$.

2. For the second problem replace cosine with its exponential form.

$\displaystyle \int_{0}^{2\pi}e^{\alpha\cos\theta}\sin(\alpha\cos \theta)d\theta$

$\displaystyle \int_{0}^{2\pi}e^{\frac{\alpha}{2}(e^{i\theta}+e^{-i\theta})}\sin(\frac{\alpha}{2}(e^{i\theta}+e^{-i\theta}))d\theta$

A substution is now in order, $\displaystyle z=e^{i\theta}$, $\displaystyle d\theta=\frac{-i dz}{z}$. This now becomes a closed contour around the unit circle in the complex plane.

$\displaystyle -i\oint_{|z|=1}\frac{e^{\frac{\alpha}{2}(z+z^{-1})}\sin(\frac{\alpha}{2}(z+z^{-1}))}{z}dz$

$\displaystyle -i\oint_{|z|=1}\frac{e^{\frac{\alpha(z^2+1)}{2z}} \sin\left(\frac{\alpha(z^2+1)}{2z}\right)}{z}dz$

Now the Cauchy integral formula can be used. However, by analyzing the original real integral, it can be shown that the integral is zero. That is the same result of the Cauchy integral formula.
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