Using $e^{ix}$ instead of sine and cosine in contour integration

Question

A while ago I asked:

Instead of using $\cos(z)$ an answerer said that is valid to use $e^{ix}$

How is this valid? I dont understand that.

score 2 · Accepted Answer · answered Dec 22 '14 at 15:32

2

$\cos z = \Re(e^{iz})$ and $\Re$ is a linear operator, so $\Re\int = \int\Re$.

answered Dec 22 '14 at 15:32

Jack D'Aurizio

Thank you Mr. Jack. But this is contour integration. Does the rule you state in the end still apply complex-contour integration? – Amad27 Dec 22 '14 at 15:34
@Amad27: yes, $\Re$ commutes with $\oint$, too. – Jack D'Aurizio Dec 22 '14 at 15:40
So we we are doing is this:
$\int_{-\infty}^{\infty} \frac{\cos(z)}{z^2 + 1} = \int_{-\infty}^{\infty} \frac{\text{Re}e^{iz}}{z^2 + 1} = \text{Re}\int_{-\infty}^{\infty} \frac{e^{iz}}{z^2 + 1}$

So if the answer is for example, $2i$ we take Re[$2i$] = 0?

just as an example, not that that is the actual value of the integral?
– Amad27 Dec 22 '14 at 15:46
@Amad27: exactly. – Jack D'Aurizio Dec 22 '14 at 15:48
@Amad27: you're welcome. – Jack D'Aurizio Dec 22 '14 at 20:49

1 Answers1