Can we find a closed form for this integral: $$\int_0^\infty e^{-ix} x^{\alpha -1} \mathrm{d}x$$
I am assuming it requires complex analysis and the use of Cauchy's residue theorem, but is there any elementary way other than complex analysis to evaluate this? I am not asking for a solution; a hint or guidance would not be overlooked. Thank you in advance!
This is the Fourier transform $\hat{f}(y)$ of $$ f(x) = x^{\alpha-1}\mathbb{1}_{\{x\geq 0\}} $$ evaluated at $y=1/(2\pi)$. If $\alpha\in(0,1)$ then $$ \begin{align*} \hat{f}(y) &= \frac{\Gamma(\alpha)}{(2i\pi y)^{\alpha}} \end{align*} $$ which implies $$ \int_0^\infty x^{\alpha-1}\,e^{-ix}\,\mathrm{d} x = i^{-\alpha}\,\Gamma(\alpha) $$
Actually, another way of saying this is $$ \Gamma(\alpha) = \int_0^{i\,\infty}z^{\alpha-1}e^{-z}\,\mathrm{d}z $$ which is a formula I already saw somewhere (but I don't know where and how one proves it usually).
To prove the formula of the Fourier transform, one can first find the dependence in $y$ by a scaling argument and then use the fact that $\langle \hat{f}|\varphi\rangle = \langle f|\hat{\varphi}\rangle$ with $\varphi$ a Gaussian (for which we know the Fourier transform) to find the right constant.
