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I am trying to figure out how to evaluate the integral $\int_{C}\frac{e^{iz}}{z(z-\pi)}dz$ where $C$ is any circle centered at the origin with radius greater than $\pi$. I can see that $\frac{e^{iz}}{z(z-\pi)}$ is analytic everywhere except where $z=0$ and $z=\pi$, both of which are in the region bounded by $C$. I can also see that by using the Taylor expansion of $e^{iz}$ that $$\int_{C}\frac{e^{iz}}{z(z-\pi)}dz = \sum_{n=0}^{\infty}\frac{i^{n}}{n!}\int_{C}\frac{z^{n-1}}{z-\pi}dz$$

I'm supposed to apply Cauchy's Theorem or Cauchy's Integral Theorem to evaluate the integral along this curve but I am not sure how to do so. I do not yet have the residue theorem in my tool box.

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  • $\begingroup$ Your question is unclear : in the title you suggest not to use the residuer theorem while in the text you seem to ask for a proof using this theorem. Could you clarify your question or change the title according to what you want, please ? $\endgroup$
    – Tom-Tom
    Mar 11, 2018 at 21:39
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    $\begingroup$ @MichaelHardy Yes sorry, I have updated the question to specify any circle centered at the origin with radius greater than $\pi$. $\endgroup$
    – Anfänger
    Mar 11, 2018 at 21:54

2 Answers 2

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Using partial fraction expansion we have, for every $R>\pi$,

$$\oint_{|z|=R} \frac{e^{iz}}{z(z-\pi)}\,dz=\frac1\pi\oint_{|z|=R} \frac{e^{iz}}{z}\,dz-\frac1\pi\oint_{|z|=R}\frac{e^{iz}}{z-\pi}\,dz$$

Now finish by using Cauchy's Integral Formula (or the residue theorem).

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  • $\begingroup$ "Hint" doesn't seem like an appropriate word for this. $\endgroup$ Mar 11, 2018 at 21:09
  • $\begingroup$ @michaelhardy Do you have a suggestion as to a more appropriate preface? $\endgroup$
    – Mark Viola
    Mar 11, 2018 at 21:12
  • $\begingroup$ I would just have omitted that word and left everything else the same. There's such a thing as a "sketch" and it seems people often use the word "hint" where "sketch" makes more sense. $\endgroup$ Mar 11, 2018 at 21:16
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    $\begingroup$ That sounds like an improvement. I've edited accordingly. $\endgroup$
    – Mark Viola
    Mar 11, 2018 at 21:19
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By symmetry the residues of $\frac{e^{iz}}{z(z-\pi)}$ at $z=0$ and $z=\pi$ are the same and they equal $-\frac{1}{\pi}$.
It follows that for any $R>\pi$ we have $$ \oint_{\|z\|=R}\frac{e^{iz}}{z(z-\pi)}\,dz = 2\pi i\cdot\left(-\frac{2}{\pi}\right) = -4i.$$

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