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Let $\gamma(t)=e^{it},t \in [0,2\pi]$. We take a look at:

$$\int_\gamma \frac{\cos(z)}{z^3+2z^2} dz$$

We let $$f(z)=\frac{\cos(z)}{z^2+2z}$$ and have

$$\int_\gamma \frac{\cos(z)}{z^3+2z^2} dz=\int_\gamma \frac{f(z)}{z}dz$$

The problem is that $\lim_{z\rightarrow 0} f(z)=\infty$. Now, I can't use the Cauchy integral formula to evaluate the integral. What can be done?

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Check your statement, you have an error –  enzotib Jul 23 '12 at 19:59
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You can instead set $f(z) = \frac{\cos(z)}{z+2}$ and use the Cauchy Integral Formula $$\frac{1}{2\pi i}\int_C \frac{f(z)}{(z-a)^{n+1}}\,dz = \frac{f^{(n)}(a)}{n!}$$ with $n = 1$. I'm pretty sure I mentioned this approach in my comment to your previous question... –  Jesse Madnick Jul 23 '12 at 20:01
    
So we have $f'(0)=-\frac{1}{4}$ and $\int_\gamma \frac{\cos(z)}{z^3+2z^2} dz=-\frac{1}{2}\pi i$. Thank you very much! –  Chris Jul 23 '12 at 20:20
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$$\frac{\cos z}{z^2(z+2)}=\frac{1}{z^2}\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...\right)\left(\frac{1}{2}\frac{1}{1+\frac{z}{2}}\right)=$$ $$\frac{1}{2z^2}\left(1-\frac{z}{2}+\frac{z^2}{4}-...\right)\left(1-\frac{z^2}{4}+...\right)=\frac{1}{2z^2}-\frac{1}{4z}+...\Longrightarrow Res_{z=0}\left(\frac{\cos z}{z^2}\right)=-\frac{1}{4}$$ Thus, the integral equals $\,\displaystyle{-\frac{1}{4}2\pi i=-\frac{\pi i}{2}}$

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