Calculation of fourier transform of a fraction How can I calculate the sine and cosine Fourier transform of the following functions?
$$f_n(t)=\frac{1}{(a^2+t^2)^n},\,n=1,2$$
What I've done so far:
For n=1:
$$\int^{\infty}_{-\infty}\frac{1}{a^2+t^2}e^{-i\omega t}sin(xt)dt=\int^{\infty}_{-\infty}\frac{1}{a^2+t^2}e^{-i\omega t}\bigg(\frac{e^{ixt}-e^{-ixt}}{2i}\bigg)dt=\int^{\infty}_{-\infty}\frac{1}{a^2+t^2}\bigg(\frac{e^{ixt-i\omega t}-e^{-ixt-i\omega t}}{2i}\bigg)dt$$
 A: If I understand the situation correctly, it suffices to evaluate $\int_{-\infty}^\infty \frac{1}{(a^2+t^2)^n} e^{-i c t} dt$ where $c$ is real but might have either sign. In the case where $c$ is positive, one should take a semicircular contour in the lower half plane, so that the real part of $-i c z$ is negative. In this case the resulting contour integral will converge. By a standard ML type estimate the contribution from the arc part of the semicircle will be zero so the desired integral is 
$$-2 \pi i \operatorname{Res} \left ( \frac{1}{(a^2+z^2)^n} e^{-i c z},z=-|a|i \right )$$
where the minus sign on the outside comes from traversing the contour clockwise. In the case $c<0$, you take the residue at $+|a|i$ and remove the minus sign. In the case $c=0$ you just have the standard arctangent integral (or, alternately, you can do the contour integration in either half plane).
A: The Fourier transform brings a Cauchy distribution into a Laplace distribution and vice-versa, hence the Fourier transform of $\frac{1}{(t^2+a^2)^n}$ is the multiple convolution of Laplace distributions.
In the integral
$$ \int_{-\infty}^{+\infty}\frac{e^{-i\omega t}}{(a^2+t^2)^n}\,dt $$
the $a$-parameter can be removed through the substitution $t\mapsto|a|z$, hence the problem boils down to finding an explicit form for
$$ I_n(\omega)=\int_{-\infty}^{+\infty}\frac{e^{-i\omega z}}{(1+z^2)^n}\,dt = \int_{-\infty}^{+\infty}\frac{e^{-i|\omega| z}}{(1+z^2)^n}\,dt$$
where the integrand function has poles of order $n$ at $z=\pm i$. By the residue theorem
$$ I_n(\omega) = 2\pi i\,\text{Res}\left(f(z),z=i\right) = 2\pi i\lim_{z\to i}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\left(\frac{e^{-i|\omega|z}}{(z+i)^n}\right) $$
so:
$$\boxed{ I_n(\omega) = \color{red}{\frac{\pi}{2^{n-1}(n-1)!} e^{-\pi|\omega|} P_{n-1}(|\omega|)} }$$
with $P_{n-1}$ being a monic polynomial with degree $n-1$:
$$ \begin{array}{|c|c|}\hline n & P_n(x) \\
\hline 0 & 1 \\
\hline 1 & x+1 \\ 
\hline 2 & x^2+3x+3 \\
\hline 3 & x^3+6x^2+15x+15 \\
\hline 4 & x^4+10x^3+45x^2+105x+105 \\ \hline \end{array} $$
