# Proving complex exponentials are orthogonal.

1. May 4, 2013

### eas123

I was wondering how you prove that ∫(e^iax)(e^ibx)dx from minus infinity to infinity is zero. When I try to evaluate this in the usual way, the result is undefined.

2. May 4, 2013

### micromass

Why do you think the result should be 0? I think that's only the case if your integral ranges from $0$ to $2\pi$ (or similar) and not on entire $\mathbb{R}$.

3. May 4, 2013

### SammyS

Staff Emeritus
Consider the following function:

$\displaystyle g(t)= \int_0^t e^{iax}e^{ibx}dx\\ \ \\ \quad\ \ \ \displaystyle =\int_0^t e^{i(a+b)x}dx$

If $\ a+b=0\,,\$ then $\ g(t)=t\ .$

So that $\lim_{t\to\,\infty}g(t)=\infty\ .$​

Otherwise, $\displaystyle \ g(t)=\frac{-i}{a+b}e^{i(a+b)t}$
$\displaystyle\quad\quad\quad\quad =\frac{1}{a+b}\left(\sin((a+b)t)-i\cos((a+b)t) \right)\ .$

So that as t→∞, g(t) oscillates with constant amplitude, 1/(a+b).​