Fourier transform of $t^n\exp(-\alpha\vert t\vert)$ $(n\geq0)$ I met this Fourier transformation when dealing with bath properties in physics. I would appreciate any clue on this.
 A: Thank @metamorphy for his suggestion. This Fourier transformation can be calculated recursively.
We first split the total integral into two parts
$$ \int_{-\infty}^{\infty} dt\ t^n \exp(-\alpha\vert t\vert-i\omega t)= \int_{0}^{\infty} dt\ t^n \exp(-\alpha t-i\omega t) + \int_{-\infty}^{0} dt\ t^n \exp(\alpha t-i\omega t).$$ Denote the first part as $I_n=\int_{0}^{\infty} dt\ t^n \exp(-\alpha t-i\omega t)$ and then take derivative with respect to $\omega$,  one can obatin
$$I_{n+1}=i\frac{\partial I_n}{\partial \omega}.$$
Starting from $I_0=\frac{1}{\alpha+i\omega}$, the recursive relation yields
$$I_n=\frac{n!}{(\alpha+i\omega)^{n+1}}.$$
The same procedure can be done for the second part. The final result reads
$$\int_{-\infty}^{\infty} dt\ t^n \exp(-\alpha\vert t\vert-i\omega t)=\frac{n!}{(\alpha+i\omega)^{n+1}}+\frac{(-1)^nn!}{(\alpha-i\omega)^{n+1}}\quad n\in \mathbb{N}$$
A: Fourier Transform of double decaying exponential with t^n polynomial growth Fourier Transform Pair
  x(t)                         X(j ω)

exp(-α|t|)                    2α/(α^2+ ω^2)

t^n x(t)                      j^n (d^n/dω^n) (X(jw))
__________________________________________________

n*************t^n x(t)***********************Xn(j ω)
0    *************   x(t)**********************2α/(ω^2+α^2)
1*******    *****   t x(t)*******************-j4ωα/(ω^2+α^2)^2
2**********          t^2 x(t)*****************(-16ω^2α)/((ω^2+α^2)^3
n********** t^n x(t)**********     (i^n)n!C(ω^n)α /((ω^2+α^2)^(n+1)
All you need for the general solution is to determine constant C.
Hint: look at the pattern generated by this table, can you see a trend is developing?

