I am struggling with taking the Fourier Transform over $t$ of the following:
$$f(t,\tau, N) = \frac{1}{1 + \big(\frac{t}{\tau}\big)^{2N}}$$
Since $\tau$ and $N$ are fixed, I can rewrite this as:
$$f(t,\tau, N) = \frac{\tau^{2N}}{\tau^{2N} + t^{2N}}$$
Letting $a = \tau^{2N}$, this gives:
$$f(t,a) = \frac{a}{a + t^{2N}}$$
So our Fourier Integral looks like:
$$ F(\omega, a) = \frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} \frac{a}{a + t^{2N}} e^{-j \omega t} dt $$
Now a handy table of Fourier pairs gives us:
$$ \frac{1}{2} e^{-a |t|} < -- > \frac{a}{a^{2} + \omega^{2}} $$
So if I multiply the integrand by $\frac{1}{a}$ I will get:
$$\frac{1}{a} \cdot \frac{a}{a + t^{2N}} = \frac{a}{a^{2} + \Big(\frac{t^{N}}{a^{\frac{1}{N}}}\Big)^{2}} $$
Does this mean:
$$ \frac{a}{a^{2} + \Big(\frac{t^{N}}{a^{\frac{1}{N}}}\Big)} <--> \frac{1}{2} e^{-a^{N} |\omega^{\frac{1}{N}}|} $$
So that:
$$ \frac{a}{a + t^{2N}} <--> \frac{a}{2} e^{-a^{N} |\omega^{\frac{1}{N}}|} $$
If not, how can I evaluate this integral? What is the fourier transform?