This is the last part of a multistage evaluation of $I(x)=\Gamma(x)\Gamma(1-x)$. Through various substitutions we get the integral $$I(x)=\int_{-\infty}^{\infty}\frac{e^{xv}}{1+e^v}dv$$
We're also given a suggested contour for this integral that is a rectangle in the complex plane with counter-clockwise path of integration from $(-R,0)$ to $(+R,0)$ on the x-axis (where $R=\infty$) then up $2\pi$ units to corner $(R,2\pi i)$ then back to the 2nd quadrant to point $(-R,2\pi i)$ and finally back to the starting point at $(-R,0)$. On the x-axis contour $z=x$ and on the upper contour we're given that $z=x+2\pi i$. We are also told to use residues.
Given this information i just let $z=v$ so $dz=dv$ and our integral becomes $$I(x)=\oint\frac{e^{xz}}{1+e^z}dz$$ We have a simple pole $z_0=\pi i$ so i evaluated $I$ as $$I(x)=2\pi i Res\left[\frac{e^{xz}}{1+e^z},z=z_0\right]=2\pi i e^{i\pi(x-1)}$$
I checked my solution by letting $x=1/2$ because we know that $I(1/2)=\Gamma(1/2)\Gamma(1-1/2)=\Gamma(1/2)\Gamma(1/2)=\pi$ But when i tried my solution i end up with $I(1/2)=2\pi$.
I'm pretty sure my problem is that i treated the $x-axis$ path as a contour by itself which enclosed the pole when it isn't a circular contour. So how do i remedy this situation? How do i include the upper half of the contour while using Residues?
Note that for $0<t<1$, we have
$$\begin{align} \lim_{R\to \infty}\oint_C \frac{e^{tz}}{1+e^z}\,dz&=\int_{-\infty}^\infty \frac{e^{tx}}{1+e^{x}}\,dx-\int_{-\infty}^\infty\frac{e^{t(x+2i\pi)}}{1+e^{x+i2\pi}}\,dx\\\\ &=\left(1-e^{i2\pi t}\right)\int_{-\infty}^\infty \frac{e^{tx}}{1+e^{x}}\,dx \tag 1\\\\ &=2\pi i \text{Res}\left(\frac{e^{tz}}{1+e^z},z=i\pi\right) \\\\ &=-2\pi i e^{i\pi t} \tag 2 \end{align}$$
Equating $(1)$ and $(2)$ yields
$$\begin{align} \int_{-\infty}^{\infty} \frac{e^{tx}}{1+e^x}\,dx&=2\pi i \frac{e^{i\pi t}}{e^{i2\pi it}-1}\\\\ &=\frac{\pi}{\sin(\pi t)} \end{align}$$