# Fourier transform of power function $t^\alpha$

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

It gives the following Fourier tranform pairs

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

Can you help with this? Thank you!

• 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 9:13

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.

• 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 3:00
• 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
• 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