Proof of $\int_0^\infty t^{a-1}e^{it}\,dt=\Gamma(a)e^{ia\pi/2}$?

Question

Can anyone show a proof of $$\int_0^\infty t^{a-1}e^{it}\,dt=\Gamma(a)e^{ia\pi/2}$$ where $0<a<1$, and $$\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt.$$ Thank you.

Answer 1

Consider an integral in the complex plane along the contour which goes from the origin out along the real axis to $+R$, then moves in a circular arc up to the point $+iR$, and then returns to the origin along the imaginary axis.