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Proving complex exponentials are orthogonal.

  1. May 4, 2013 #1
    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.

    Thanks in advance for your help!
     
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  3. May 4, 2013 #2

    micromass

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    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}##.
     
  4. May 4, 2013 #3

    SammyS

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    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).​
     
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