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!
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.