I was trying to obtain Schröder's integral formula for the Gregory coefficients.
The starting point was another integral representation:
$$G_n = \int_0^1 \binom{x}{n} dx$$
$$\binom{x}{n}=\frac{\Gamma(1+x)}{n! \Gamma(1-n+x)}= \frac{1}{\pi n!} \sin ( \pi x) \Gamma(n-x) \Gamma(1+x)= \\ = \frac{1}{\pi} \sin ( \pi x) B (n-x,1+x) = \frac{1}{\pi} \sin ( \pi x) \int_0^1 t^{n-x-1} (1-t)^x dt$$
$$\frac{1}{\pi} \int_0^1 \sin ( \pi x) \frac{(1-t)^x}{t^x} dx= \frac{1+ \frac{1}{t}-1}{\pi^2+\log^2 \left(\frac{1}{t}-1 \right)}$$
$$G_n=\int_0^1 \frac{t^{n-2}}{\pi^2+\log^2 \left(\frac{1}{t}-1 \right)} dt=$$
$$=\int_1^\infty \frac{t^{-n}}{\pi^2+\log^2 \left(t-1 \right)} dt=\int_0^\infty \frac{dt}{(1+t)^n \left(\pi^2+\log^2 t \right)}$$
Everything seems fine, except I lost the sign somewhere, because it should be:
$$G_n=(-1)^{n+1} \int_0^\infty \frac{dt}{(1+t)^n \left(\pi^2+\log^2 t \right)}$$
Where's my mistake?
You made a mistake when applying the Gamma reflection formula with $z=x+1-n$: $$\Gamma(z)=\frac{\pi}{\sin(\pi z)}\frac{1}{\Gamma(1-z)}$$ So in fact, the correct step should be: $$\frac{\Gamma(1+x)}{n! \Gamma(x+1-n)}=\frac{1}{\pi n!}\color{red}{\sin(\pi(x+1-n))}\Gamma(n-x)\Gamma(1+x)$$ To which the part in red simplifies to (if $n$ is an integer): $$\sin(\pi(x+1-n))=(-1)^{n+1}\sin(\pi x)$$