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 $a$?
 A: To me the most obvious action would be:
$$\int_{-\infty}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)} dx=\frac{1}{\Gamma(a)}\int_{-\infty}^\infty B(x,a)\,\sin(\pi x) dx$$
$$\int_{-\infty}^\infty B(x,a)\,\sin(\pi x) dx=\int_{-\infty}^\infty \int_0^1 t^{x-1}(1-t)^{a-1} \sin(\pi x) ~dt~ dx$$
However the integral for $x$ diverges, we can only find the part for $x>0$ (because $\ln t<0$):
$$\int_{0}^\infty t^{x} \sin(\pi x) ~dx=\int_{0}^\infty e^{(\ln t)x} \sin(\pi x) ~dx= \frac{\pi}{\pi^2+\ln^2 t}$$
So we have for the part with $x>0$:
$$\int_{0}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)} dx=\frac{\pi}{\Gamma(a)} \int_0^1 \frac{(1-t)^{a-1}}{t(\pi^2+\ln^2 t)} ~dt$$
We can make an obvious substitution:
$$ \int_0^1 \frac{(1-t)^{a-1}}{t(\pi^2+\ln^2 t)} ~dt=\int_0^\infty \frac{(1-e^{-u})^{a-1}}{\pi^2+u^2} du$$
Or we can use the integral representation of the Gregory coefficients somehow (thanks, Jack D'Aurizio!):
$$\displaystyle G_n=(-1)^{n-1}\!\int\limits_0^{\infty}\frac{dx}{(1+x)^n\left(\ln^2 x + \pi^2\right)}$$
If this leads anywhere, I'll edit my post.
A: Interesting question. By using the reflection formula, your integral can be written as a convolution integral:
$$ I(a) = \pi \int_{-\infty}^{+\infty}\frac{dx}{\Gamma(x+a)\Gamma(1-x)}.\tag{1}$$
We may also notice that when $n\in\mathbb{N}$ we have:
$$ \int_{-\infty}^{+\infty}\frac{\sin(\pi x)\,dx}{x(x+1)\cdot\ldots\cdot(x+n)}=\pi\sum_{k=0}^{n}(-1)^k \text{Res}\left(\frac{1}{x(x+1)\cdot\ldots\cdot(x+n)},x=-k\right)\tag{2}$$
where the $RHS$ of $(2)$ equals:
$$ \frac{\pi}{n!}\sum_{k=0}^{n}\binom{n}{k} = \color{red}{\frac{\pi\, 2^n}{n!}}\tag{3} $$
so the most reasonable conjecture is:

$$ I(a) = \pi \int_{-\infty}^{+\infty}\frac{dx}{\Gamma(x+a)\Gamma(1-x)}\stackrel{!}{=}\color{red}{\frac{\pi\, 2^{a-1}}{\Gamma(a)}}.\tag{4}$$

Numerical simulations support $(4)$. Probably it is enough to exploit the log-convexity of the $\Gamma$ function to extend the validity of $(4)$ from $a\in\mathbb{N}$ to $a\in\mathbb{R}^+$.
Update: As a matter of fact, this Sangchul Lee's blogpost proves that the partial fraction decomposition through the residue theorem still works for $a\in\mathbb{R}^+\setminus\mathbb{N}$, setting $(4)$ as an identity. Credits to him for another great piece of cooperative math.
Following Yuriy S' approach,
$$ \int_{-\infty}^{+\infty}B(a,x)\sin(\pi x)\,dx \\=\int_{0}^{+\infty}\int_{0}^{1}\left[(1-t)^{a-1}t^{x-1}\sin(\pi x)+t^{x-a}(1-t)^{a-1}\sin(\pi(a-x))\right]\,dt\,dx$$
leads to an integral that can be easily evaluated through the residue theorem, giving an alterntive proof of $(4)$.
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 $a>0$.
