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Let $\gamma\colon [0,2\pi]\longrightarrow \mathbb{C}, t\mapsto e^{it}$ be a contour.

I want to compute $\int_\gamma \cos(z)/z^3\mathrm{d}z$ by using the polynomial series for $\cos(z)$.

So I get $$\int_\gamma \sum_{n=0}^\infty (-1)^n \frac{z^{2n-3}}{(2n)!}\mathrm{d}z$$

Hence, by evaluating on the curve: $$\int_{0}^{2\pi}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} i\exp\big(2it(n-1)\big)\mathrm{d}t$$

I am stuck here.

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3 Answers 3

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Hint: integrate your sum term by term. If you want to compute it in a more direct way, write $\cos(z) = \frac{1}{2}\left( e^{iz} + e^{-iz}\right)$ and integrate directly. For yet another way, utilize Cauchy's integral formula.

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  • $\begingroup$ If I do it term by term, I constantly yield zero, as I am left with somewhat like this $\sim \int^{2\pi}_0 e^{it(2n-2)}\mathrm{d}t$. $\endgroup$ Commented Nov 15, 2017 at 14:56
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    $\begingroup$ @EpsilonDelta, All of the terms except for one of them are identically equal to zero. Try to see which one that is. $\endgroup$
    – Marcus M
    Commented Nov 15, 2017 at 15:01
  • $\begingroup$ Oh, you are correct, I completely missed it. The $n=1$ term does not cancel. My result would now be $-\pi i$ for the whole integral, is this correct? $\endgroup$ Commented Nov 15, 2017 at 15:10
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    $\begingroup$ @EpsilonDelta I think you forgot about the $i$ hanging around; you should get $- \pi i$ $\endgroup$
    – Marcus M
    Commented Nov 15, 2017 at 15:13
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$$\int_\gamma \frac{\cos z}{z^3}\,\mathrm{d}z=2\pi i\,\text{Res}\left(\frac{\cos z}{z^3},\{z,\;0\}\right)=2\pi i\,\left(-\frac{1}{2}\right)=-i \pi$$

Hope this helps

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It is easy to see that $z=0$ is the only singularity in the unit disk, $|z|=1$.

An easier approach is to use the Cauchy Integral Formula For Derivatives: $$\int_{\gamma} \dfrac{f(z)}{(z-a)^{m+1}}dz=2\pi i .\dfrac{f^{(m)}(a)}{m!}$$ for $m>1.$

In this case, $f(z)=\cos z$, $a=0$ and $m=2$.

Plugging in the formula, $$\int_{\gamma} \dfrac{\cos z}{z^{3}}dz=\dfrac{2 \pi i}{2!}.-\cos (0)=-\pi i.$$

All the best.

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