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While studying the 1/f noise, I found this webpage http://www.dsprelated.com/showarticle/40.php

It gives the following Fourier tranform pairs

enter image description here

However, there are no detailed explanation on how this formula is derived.

Can you help with this? Thank you!

ecook
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  • Note that here $u(t)$ is the unit step function. Thus the Fourier transform reduces to the Laplace transform which is nearly the definition of the gamma function. – Urgje Sep 26 '14 at 09:13

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Gamma function is defined by $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx$$

Now set $z=\alpha+1$, $x=t$, and you get this:

$$\Gamma(\alpha+1)=\int_0^\infty t^{\alpha}e^{-t}dt.$$

And Fourier transform is defined by

$$\hat f(\omega)=\int_{-\infty}^\infty f(t)e^{-i \omega t}dt.$$

Substituting $f(t)=u(t)t^\alpha$, you get:

$$\hat f(\omega)=\int_0^\infty t^\alpha e^{-i\omega t}dt.$$

This is almost the above formula for Gamma function. Now substitute $v=i\omega t$, then $t=\frac v{i\omega}$, and we have:

$$\hat f(\omega)=(i\omega)^{-\alpha-1}\int_0^\infty v^\alpha e^{-v}dv=(i\omega)^{-\alpha-1}\Gamma(\alpha+1).$$

Now it's the matter of taking absolute value and argument of the answer to compute the magnitude and phase.

Ruslan
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    Thank you for the details. I have one question. The substitution $\nu=i\omega t$ you give maps t on interval $(0,+\infty)$ to $\nu$ on interval $(0,i\infty)$. In other words, the integration range is changed from real axis to imaginary axis. Does the definition of Gamma function still hold when the integration range is imaginary? – ecook Sep 26 '14 at 10:10
  • Sorry, I can't really explain this part because of somewhat lacking knowledge (although I suspected it's not quite fair to omit it), but when I give this integral to Mathematica, it says that it's equal to $\Gamma(\alpha+1)$ provided $-1<\Re\alpha<0$. I hope someone here could explain this missing part better. Also I think this limitation on $\alpha$ is stronger than actual one, my method of showing the result may be too restrictive. – Ruslan Sep 26 '14 at 10:35
  • Anyway, thank you for your explanation – ecook Sep 27 '14 at 03:00
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    Using (the idea in the proof of) Jordan's lemma it's possible to show that integral along quarter circle of radius $R$ from real axis to imaginary axis tends to zero as $R \to \infty$ as long as $\alpha < 0$. Since the integrand has no residues inside the closed curve along real axis to $R$, the quarter circle, the imaginary axis from $iR$ to $0$ this gives the integral zero along the closed contour and thus the integral along real axis and imaginary axis are equal in the limit $R \to \infty$. The condition $\alpha > -1$ comes from convergence of the integral near the endpoint 0. – dioid Sep 27 '14 at 15:09
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    Another difficulty with the proof is that the integral formula defining the Fourier transform is only valid for $f \in L^1$ but $u(t)t^\alpha \not \in L^1$ for any $\alpha$. This can be handled by extending the domain of Fourier transform by continuity to $L^1+L^2$ and using density of $C_0^\infty$ into interpreting the integral as an improper integral with limit in $L^2$ sense, at least for $-1 < \alpha \leq -\frac{1}{2}$. For $\alpha > -\frac{1}{2}$ the Fourier transform has to be interpreted in the sense of tempered distributions. – dioid Sep 27 '14 at 15:33
  • @dioid very nice explanation. Thank you – ecook Oct 02 '14 at 12:37