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

Question

I am trying to prove the identity $$\Gamma(z) e^{i \pi z/2} = \int_0^\infty t^{z-1} e^{it}\, dt$$ for $0 < \Re(z) < 1$, starting from the integral definition of the gamma function $$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\, dt.$$ I've been trying things for a while and don't really know how to proceed. I've tried a couple of tricks involving taking complex conjugates of both sides of the identity, and trying to relate those back to the original gamma function in some enlightening way. I've also tried thinking of a clever variable substitution for $z$, or even arguing by power series, though I don't know how I would handle the difficulties of convergence. Any helpful pointers would be appreciated.

Answer 1

Let $D=\left\{z=re^{i\theta}\in\mathbb C|-\frac{3\pi}{4}\leq\theta\leq\frac{3\pi}{4}\right\}$, and consider the function $f(w)=w^{z-1}e^{-w}$ on $D$, where we choose the $\mathrm{exp}((z-1)(\log|w|+i\mathrm{Arg} w))$ branch of $w^{z-1}$. Let $\varepsilon,M>0$ with $\varepsilon <M$, and consider the following contour: $$\gamma_1(t)=t,\,\,t\in[\varepsilon, M]\\\gamma_2(t)=Me^{it},\,\,t\in\left[0,-\frac{\pi}{2}\right]\\\gamma_3(t)=ti,\,\,t\in[-M,-\varepsilon]\\\gamma_4(t)=\varepsilon e^{it},\,\,t\in\left[-\frac{\pi}{2},0\right].$$ $f$ is holomorphic in $D$, therefore $$\int_{\gamma_1}f+\int_{\gamma_2}f=\int_{-\gamma_3}f+\int_{-\gamma_4}f.$$ Then, prove that the second and fourth integral converge to $0$ (for the second you will need that the real part of $z$ is less than $1$, for the fourth that it's greater than $0$), the first integral converges to $\Gamma(z)$, while the third converges to $$e^{-\pi iz/2}\int_0^{\infty}t^{z-1}e^{it}\,dt.$$