I am currently looking at an example of how to calculate the Fourier Transform for the function \begin{equation} f(x) = \exp\left({-\frac{x^2}{2}}\right) \end{equation} Now $f$ solves the differential equation \begin{equation} f'(x) = -xf(x) \end{equation} and so, applying the FT to both sides gives \begin{equation} i \xi\,\hat{f}(\xi) = -i(\hat{f})'(\xi) \end{equation} Using these two equations we can derive that \begin{equation} \left(\frac{\hat{f}(x)}{f(x)}\right)' = 0 \end{equation} hence \begin{equation} \hat{f}(x) = c_0f(x) = c_0\exp\left({-\frac{x^2}{2}}\right) \end{equation}

And here is where I have trouble, because the author of the notes that I am using says that $c_0 \geq 0$, which I can't see why .. I am sure it must be something obvious and I am just too blind to see it. Any hint would be highly appreciated, many thanks !

Edit: It is not the claim that I don't understand (I know it is right), it is just that I can't see why this is true looking solely at the derivations I have so far.

  • $\begingroup$ Compare the last to first line in your question. You sure you want to revert the $x^2$ I revised in? $\endgroup$ – anon Jan 31 '12 at 2:05
  • $\begingroup$ @anon: thks for mentioning this, was a typo. $\endgroup$ – harlekin Jan 31 '12 at 8:45
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    $\begingroup$ BTW, why bother to look at the quotient (i.e. $(\frac{\hat{f}}{f})'$), you see $f$ and $\hat{f}$ satisfy the same ODE, and hence they differ by a constant. $\endgroup$ – AD. Feb 4 '12 at 13:28


  • $\begingroup$ ok thks! I calculated that as well so I know the author is correct, however, he then goes on and does something slightly different to determine the actual value of $c_0$. Hence I was wondering whether I am actually missing something within the lines I gave above that should tell me $c_0$ is non-negative .. $\endgroup$ – harlekin Jan 31 '12 at 1:05
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    $\begingroup$ @harlekin: Without the notes in front of me it's impossible to tell. $\endgroup$ – anon Jan 31 '12 at 1:07
  • $\begingroup$ hm .. I understand, thanks for your answer anyways ! $\endgroup$ – harlekin Jan 31 '12 at 1:19
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    $\begingroup$ @harlekin: even without the final "$=\sqrt{2\pi}$" in anon's computation (which is not obvious enough to be left to the reader) it should be clear that the integral is at least positive because the integrand is positive everywhere. $\endgroup$ – hmakholm left over Monica Jan 31 '12 at 1:21
  • $\begingroup$ yes that's right! $\endgroup$ – harlekin Jan 31 '12 at 1:48

You might want to have another proof. Well, it does not hurt.

We can define, for any complex $\xi$,

$$\hat{f} (\xi) = \int_{-\infty}^{+\infty} e^{-i x \xi} e^{-\frac{x^2}{2}} dx.$$

This is well-defined, and you can check that $\hat{f}$ is analytic. There are some general results that you can apply to prove it; I don't remember them precisely, but with a gaussian, everything works well. Now, if we compute $\hat{f}$ for imaginary numbers, we have:

$$\hat{f} (iy) = \int_{-\infty}^{+\infty} e^{xy} e^{-\frac{x^2}{2}} dx = \int_{-\infty}^{+\infty} e^{\frac{y^2}{2}} e^{-\frac{(x-y)^2}{2}} dx = e^{\frac{y^2}{2}} \int_{-\infty}^{+\infty} e^{-\frac{x^2}{2}} dx = \sqrt{2\pi} e^{-\frac{(iy)^2}{2}}.$$

Hence, $\hat{f} (\xi) = \sqrt{2\pi} e^{-\frac{\xi^2}{2}}$ for imaginary $\xi$. By analytic continuation, this identity holds for any $\xi$.


Note that inverse Fourier transform is defined as $$\hat{f}(\zeta) =\int_{-\infty}^{\infty}f(x)e^{-j\zeta x}dx$$ From your solution we can see that $$c_0 = \hat{f}(0) = \int_{-\infty}^{\infty}f(x)dx$$ which is nothing but the area of the Gaussian which is always positive...



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