I want to solve the following integral:
$$ I(t) = \int_{0}^{\infty} \frac{1}{2^{k/2} \Gamma(\frac{k}{2})} x^{\frac{k}{2}-1}e^{-\frac{x}{2}} e^{jxt} dx $$
Where $j = \sqrt{-1}$, and $k \in \mathbb{N}$ and even.
I know that the answer is:
$$ I(t) = (1-j2t)^{-\frac{k}{2}}$$
So far, I let:
$$ u = \frac{x}{2} \Rightarrow x=2u, dx=2du $$
Thus we have:
$$ I(t) = \int_{0}^{\infty} \frac{1}{2^{k/2} \Gamma(\frac{k}{2})} (2u)^{\frac{k}{2}-1}e^{-u} e^{j2ut} 2du $$ $$ I(t) = \frac{1}{\Gamma(\frac{k}{2})} \int_{0}^{\infty} 2^{-k/2} 2^{k/2}u^{\frac{k}{2}-1}e^{ju(2t+j)} du $$ $$ I(t) = \frac{1}{\Gamma(\frac{k}{2})} \int_{0}^{\infty} u^{\frac{k}{2}-1}e^{ju(2t+j)} du $$
Now I let:
$$ a = \frac{k}{2}-1 \quad b = 1-2t$$
This gives us:
$$ I(t) = \frac{1}{\Gamma(a+1)} \int_{0}^{\infty} u^{a}e^{-jbu} du $$
Now I evaluated this via integration by parts. Letting:
$$ f(u) = u^{a} \quad g(u) = e^{-jbu} $$
We have
$$ I(t) = \frac{1}{\Gamma(a+1)} \Big(\sum_{n=0}^{\infty} (-1)^{n} (f^{(n)} \cdot g^{[n+1]})(u)\Big) \Big|_{0}^{\infty}$$
Where $f^{(n)}$ is the $n$-th derivative of $f$ and $g^{[n+1]}$ is the ($n+1$)-st antiderivative of $g$. This evaluates to:
$$ I(t) = \Big(\sum_{n=0}^{a} (-1)^{n} \frac{u^{a-n}}{(a-n)!} (-jb)^{-(a+1)} e^{-jbu} \Big)\Big|_{0}^{\infty}$$
However this is where I am stuck. I am having trouble taking the limits -- I just get infinity and zero.
What should I do next? Is this even the right approach? Any other ideas?