For $a>0$ i have managed to show that this is the Fourier transform of the function.

$$ \mathcal{F}[e^{-a|x|}](s) = \frac {2a}{\sqrt{2{\pi}}(a^2+s^2)}. $$

How do I now use this to find the Fourier transform of:

$$ \mathcal{F}[\log(a^2+s^2)](s)? $$

I have tried to apply the inversion formula for Fourier transforms but I haven't had any success. Any help would be greatly appreciated.

  • $\begingroup$ What is the derivative of $\log(a^2+s^2)$ with respect to $s$? What do you know about the Fourier transform of a primitive? $\endgroup$ – Jack D'Aurizio May 13 '15 at 18:33
  • $\begingroup$ so we have $\frac{d}{ds}(log(a^2+s^2)=\frac{2s}{a^2+s^2}$ but im not sure what you mean for the next part $\endgroup$ – sean May 13 '15 at 18:42
  • $\begingroup$ any chance you could just show me how to do it please? kinda desperate my exam is tomorrow $\endgroup$ – sean May 13 '15 at 18:55
  • $\begingroup$ Do you know the formula for the Fourier transform of the derivatives of a function? $\endgroup$ – Lost May 13 '15 at 18:57
  • $\begingroup$ unfortunately not. $\endgroup$ – sean May 13 '15 at 18:58

The derivative of $\log(a^2+s^2)$ is $\frac{2s}{a^2+s^2}$. Now, we know that the fourier transform of $f'(x)$ is $i\omega \hat{f}(\omega)$. So we can find the fourier transform of $\frac{2s}{a^2+s^2}$ and then divide it by $i\omega$ to find our answer. We can use the Residue Theorem to evaluate $\int_{-\infty}^{\infty} \frac{2s}{a^2+s^2}e^{-i\omega s}ds$.

There are poles at $\pm ia$. Depending on the sign of $\omega$, we either do our contour integral in the upper half or lower half of the plane. Let's do it in the lower half plane. The only residue in our domain is at $-ia$. Our residue is :$$\frac{2s}{2s}e^{-i \omega s}=e^{-i \omega s}$$ For $s=-ia$, we get $e^{-\omega a}$

With the Residue Theorem, we get $-2 \pi i e^{-\omega a}$ (minus because we are in the lower half plane), which is the Fourier Transform of the derivative. Now we divide by $i \omega$ to find the FT of the original function, and we get : $$\frac{-2 \pi}{\omega}e^{- \omega a}$$ which we could rewrite as ($\omega= 2 \pi \xi $ , $\xi$ is the frequency) $$\frac{-1}{\xi}e^{- \omega a}$$

Hope this helps !

Note 1) : that this question was never answered for 3 years is not a reason to never answer it.
Note 2) : this is my first answer. If I did any mistake, please say it in the comments and I will correct my answer. Thank you !


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