1. ## Integral of e

Hi,

I've to show that

$\int_0^{2\pi}e^{-i(k+n)\theta}d\theta=0$

Thanks for any help.

2. Originally Posted by Arczi1984
Hi,

I've to show that

$\int_0^{2\pi}e^{-i(k+n)\theta}d\theta=0$

Thanks for any help.
$\int_0^{2\pi}{e^{-i(k + n)\theta}\,d\theta} = -\frac{1}{i(k + n)}\left[e^{-i(k + n)\theta}\right]_0^{2\pi}$

$= -\frac{1}{i(k + n)}\left[e^{-2\pi i(k + n)} - e^{0i(k + n)}\right]$.

Assuming that $-(k + n)$ is an integer, these two terms are actually equal (since an integer multiple of $2\pi$ represents a number of revolutions about the unit circle back to an angle of $0$).

$= -\frac{1}{i(k + n)}\left[1 - 1\right]$

$= -\frac{1}{i(k + n)}\left[0\right]$

$= 0$.

3. Originally Posted by Prove It
$\int_0^{2\pi}{e^{-i(k + n)\theta}\,d\theta} = -\frac{1}{i(k + n)}\left[e^{-i(k + n)\theta}\right]_0^{2\pi}$

$= -\frac{1}{i(k + n)}\left[e^{-2\pi i(k + n)} - e^{0i(k + n)}\right]$.

Assuming that $-(k + n)$ is an integer, these two terms are actually equal (since an integer multiple of $2\pi$ represents a number of revolutions about the unit circle back to an angle of $0$).

$= -\frac{1}{i(k + n)}\left[1 - 1\right]$

$= -\frac{1}{i(k + n)}\left[0\right]$

$= 0$.
Be careful if $k = -n$.