6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '18 at 21:39
  • 1
    $\begingroup$ @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 '18 at 21:54
6
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ "Hint" doesn't seem like an appropriate word for this. $\endgroup$ – Michael Hardy Mar 11 '18 at 21:09
  • $\begingroup$ @michaelhardy Do you have a suggestion as to a more appropriate preface? $\endgroup$ – Mark Viola Mar 11 '18 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. $\endgroup$ – Michael Hardy Mar 11 '18 at 21:16
  • 1
    $\begingroup$ That sounds like an improvement. I've edited accordingly. $\endgroup$ – Mark Viola Mar 11 '18 at 21:19
0
$\begingroup$

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 $$ \oint_{\|z\|=R}\frac{e^{iz}}{z(z-\pi)}\,dz = 2\pi i\cdot\left(-\frac{2}{\pi}\right) = -4i.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.