Integral $\int_{-\infty}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)}\,dx$ I would like to evaluate this integral:
$$\mathcal F(a)=\int_{-\infty}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)}\,dx,\quad a>0.\tag1$$
For all $a>0$ the integrand is a smooth oscillating function decaying for $x\to\pm\infty$. The poles of the gamma function in the numerator are cancelled by the sine factor.
For $a\in\mathbb N$, the ratio of the gamma functions simplifies to a polynomial in the denominator, and in each case the integral can be pretty easily evaluated in a closed form, e.g.
$$\mathcal F(3)=\int_{-\infty}^\infty\frac{\sin(\pi x)}{x\,(x+1)\,(x+2)}\,dx=2\pi.\tag2$$
Can we find a general formula for $\mathcal F(a)$ valid both for integer and non-integers positive values of $a$?

Let
$$g(x) = \frac{\Gamma(x)\,\exp(i\pi x)}{\Gamma\left(x+a\right)},\quad a>0$$
and
$$\mathcal G(a)=\int_{-\infty}^\infty g(x)\,dx.$$
The residues of $g$ are along the real line at $z\in-\mathbb N$:
\begin{align}
r_n&=\operatorname*{Res}_{z = -n}g(z) \\
&= \lim_{z\to-n}  (z+n) \frac{\Gamma(z+n+1)}{(z+n)(z+n-1)\cdots(z+1)} \frac{e^{i\pi z}}{\Gamma\left(z+a\right)}\\
&= \frac{e^{-i\pi n}}{\Gamma(a-n)} \frac{(-1)^n}{n!} \\
&= \frac{1}{n! \Gamma(a-n)}
\end{align}
Then, considering a semicircular contour along $\mathbb R$ and $R e^{i\theta}$, indented around each pole, we have
$$\mathcal G(a) = \pi i\sum_{n=0}^\infty r_n = \pi i \frac{1}{\Gamma(a)}\sum_{n=0}^\infty {a-1\choose n} = \pi i \frac{2^{a-1}}{\Gamma(a)}.$$
It is clear that the outer part of the contour makes no contribution since for large $|z|$, $|g(z)| \le \frac{R e^{-\pi y}}{|P_{\lfloor a \rfloor}(z)|}$.
Now take the complex part to conclude
$$\boxed{\mathcal F(a) = \pi\frac{2^{a-1}}{\Gamma(a)}}$$
which is valid for all $a>0$.
