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 $a$?

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 $(4)$.
