Representation of Gamma function on $0 < \operatorname{Re } z < 1$ I've read that for $0 < \operatorname{Re }z < 1$ the Gamma function has the following representation:
$$\Gamma(z) = e^{i\pi \frac{z}{2}} \int_0^\infty t^{z-1} e^{-it} \, dt.$$
I couldn't find any proof of this, does someone have a reference? Your help is appreciated.
 A: For $R > r>0$ consider the contour consisting of the four paths
\begin{align}
&\gamma_1 \colon [r,R] \to \mathbb{C} \, , \, t \mapsto t \, , \\
&\gamma_2 \colon \left[0,\frac{\pi}{2}\right] \to \mathbb{C} \, , \, \phi \mapsto R \mathrm{e}^{\mathrm{i} \phi} \, , \\
&\gamma_3 \colon [r,R] \to \mathbb{C} \, , \, s \mapsto \mathrm{i} (R + r - s) \, , \\
&\gamma_4 \colon \left[0,\frac{\pi}{2}\right] \to \mathbb{C} \, , \, \phi \mapsto r \mathrm{e}^{\mathrm{i} \left(\frac{\pi}{2} - \phi \right)} \, . \\
\end{align}
For $z \in \mathbb{C}$ with $\operatorname{Re} (z) \in (0,1)$ the function 
$$ f \colon \mathbb{C}\setminus \mathbb{R}_{\leq 0} \to \mathbb{C} \, , \, \eta \mapsto \eta^{z-1} \mathrm{e}^{- \eta} \, ,$$
is holomorphic, so 
$$ \sum \limits_{k=1}^4 \int \limits_{\gamma_k} f(\eta) \, \mathrm{d} \eta = 0 $$
holds by Cauchy's integral theorem. We have
\begin{align}
\left| \, \int \limits_{\gamma_2} f(\eta) \, \mathrm{d} \eta \,\right| &= \left| \,\int \limits_0^{\pi/2} (R \mathrm{e}^{\mathrm{i} \phi})^z \mathrm{e}^{-R \mathrm{e}^{\mathrm{i} \phi}} \mathrm{i} \, \mathrm{d} \phi \, \right| \leq R^{\operatorname{Re}(z)} \mathrm{e}^{\pi | \operatorname{Im}(z)| /2} \int \limits_0^{\pi/2} \mathrm{e}^{- R \cos(\phi)} \, \mathrm{d} \phi \\
&\leq R^{\operatorname{Re}(z)} \mathrm{e}^{\pi | \operatorname{Im}(z)| /2} \int \limits_0^{\pi/2} \mathrm{e}^{- R \left(1-\frac{2}{\pi} \phi\right)} \, \mathrm{d} \phi = \frac{\pi}{2} \mathrm{e}^{\pi | \operatorname{Im}(z)| /2} (1 - \mathrm{e}^{-R}) R^{-(1-\operatorname{Re}(z))} \stackrel{R\to \infty}{\longrightarrow} 0
\end{align}
and, similarly, 
$$ \left| \, \int \limits_{\gamma_4} f(\eta) \, \mathrm{d} \eta \,\right| \leq \frac{\pi}{2} \mathrm{e}^{\pi | \operatorname{Im}(z)| /2} (1 - \mathrm{e}^{-r}) r^{-(1-\operatorname{Re}(z))} \stackrel{r\to 0}{\longrightarrow} 0 \, .$$
Therefore
\begin{align}
\Gamma(z) &= \int \limits_0^\infty f(t) \, \mathrm{d} t = \lim_{R \to \infty} \lim_{r \to 0} \int \limits_{\gamma_1} f(\eta) \, \mathrm{d} \eta = - \lim_{R \to \infty} \lim_{r \to 0} \int \limits_{\gamma_3} f(\eta) \, \mathrm{d} \eta \\
&= - \lim_{R \to \infty} \lim_{r \to 0} \int \limits_r^R [\mathrm{i} (R+r-s)]^{z-1} \mathrm{e}^{- \mathrm{i} (R+r-s)} (- \mathrm{i}) \, \mathrm{d} s \\
&= i^{z} \lim_{R \to \infty} \lim_{r \to 0} \int \limits_r^R t^{z-1} \mathrm{e}^{-\mathrm{i} t} \, \mathrm{d} t = \mathrm{e}^{\mathrm{i} \frac{\pi}{2} z} \int \limits_0^\infty t^{z-1} \mathrm{e}^{-\mathrm{i} t} \, \mathrm{d} t \, .
\end{align}
