I am revisiting the Fourier transform and I found great lecture notes by Professor Osgood from Standford (pdf ~30MB).

On page 30 and 31 he show that the complex exponentials form an orthonormal basis. I understand the result, but not his calculation. He shows that the inner product of two different exponentials $(e_n (t) = e^{2\pi int}, e_m(t)= e^{2\pi imt})$ with $m \neq n$ is $0$ (He uses round parenthesis to denote the inner product). So, he does the calculation:

$$ (e_n, e_m) = \int_0^1 e^{2\pi int} \overline{e^{2\pi imt}}dt = \dots = \frac{1}{2\pi i(n -m)} (e^{2\pi i(n - m)} - e^0) = \frac{1}{2\pi i(n -m)} (1 - 1) = 0$$

So why is $e^{2\pi i(n - m)} = 1$? And why do I have to look at the case $ m = n$ separately and cannot also just plug it into the last step? (I am aware that I would not get a sensible result then, but it still seems strange). Does this have anything to do because I am using Lebesgue integration and complex numbers? I suppose I have to review some math basics ...

  • $\begingroup$ What is the anti-derivative of $e^{2\pi i (n-m)t}$? Does the formula hold for $n = m$? $\endgroup$ – Christopher A. Wong Jun 8 '14 at 19:16
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    $\begingroup$ $e^{i\theta} = \cos \theta + i\sin \theta$ $\endgroup$ – Paul Hurst Jun 8 '14 at 19:16
  • $\begingroup$ Ok, so if understand you correctly it's much easier. They become zero for all exponents because of the 2 pi and the integral for m=n is not defined because otherwise the denominator isn't defined anymore? $\endgroup$ – Lucas Jun 8 '14 at 19:50
  • $\begingroup$ The Integral is defined. But the antiderivative of $e^{2\pi i (n-m) x} = 1$ is the identity, so that the definite (normalized) integral is $1$ instead of $0$. $\endgroup$ – PhoemueX Jun 8 '14 at 20:27
  • $\begingroup$ You seems to have some trouble with the exponential function. For a crash course on it I recommend reading the introduction to Rudin's Real and complex analysis. This Q&A, more advanced, might also be interesting. $\endgroup$ – Giuseppe Negro Jun 8 '14 at 20:38

If $l=n-m\neq 0$ you have

$$e^{2\pi i(n-m)}=e^{2\pi i l}=\cos(2\pi l)+i\sin(2\pi l)=1+i\cdot 0=1$$

The case $m=n$ has to be considered separately because in the solution you showed you divide by $m-n$, which you can't do if $m-n=0$. For $m=n$ you simply get

$$\int_0^1e^{2\pi i nt}e^{-2\pi i nt}dt=\int_0^1dt=1$$

  • $\begingroup$ I know that the integral must be over $[0,1)$ in order to work, but do you know why is that? Is it because the circles have radius of 1? $\endgroup$ – Ooker Nov 6 '17 at 20:13

You can of course consider $m=n$ case separately, as Matt suggested.

Alternatively, you can use L'Hospitals rule, in the expression you found, and see that in the limit $m \to n,$ the integral indeed converges to $1.$


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