How can I prove that the Fourier transform of $\log(1 + x^2)$ is given by $-\frac{\sqrt{2\xi} e^{-|\xi|}}{|\xi|}$? (I am using $\xi$ as my Fourier variable.)
Fourier Transform log(1 + x^2)
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fourier-analysis
fourier-transform
1 Answers
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You have for some constant $c$ that the Fourier Transform (w.r.t. $x$) of $f(c,x) = 1/(x^2 + c)$ is given by $F(c,\xi) = \sqrt{\frac{\pi}{2 c}} \exp(-\sqrt c|\xi|)$.
[You can find this result e.g. in Wikipedia in the table "Quadratisch integrierbare Funktionen" where you set $c = a^2$.]
Now regarding $c$ as a variable and integrating w.r.t. $c$ gives $g(c,x) = \int f(c,x) \rm{d} c = \log (x^2 + c)$. Hence the Fourier Transform of $g(c,x)$ is (apart from some possible constant) $$G(c,x) = \int F(c,x) {\rm{d}} c = \int \sqrt{\frac{\pi}{2 c}} \exp(-\sqrt c|\xi|) {\rm{d}} c \quad .$$
Substituting $c= b^2$ gives ${\rm{d}} c = 2 b \, {\rm{d}} b $ and
$$G(b,x) = \int \sqrt{2 {\pi}} \exp(- b |\xi|) \rm{d} b = - \sqrt{2 {\pi}} \frac{1}{|\xi|} \exp(- b |\xi|)\quad .$$
Now setting $c=1$ (which corresponds to $b=1$) gives $g(1,x) = \log (x^2 + 1)$ which is your function. Hence the Fourier Transform of that function is $G(1,x) = - \sqrt{2 {\pi}} \frac{1}{|\xi|} \exp(- |\xi|)\;$. So the result that you claim in your question is not quite right: possibly just a typo where you set \xi instead of \pi .