complex analysis - Analytic continuation of Euler's reflection formula with the Gamma function

Let $\widetilde\Gamma$ be an analytic continuation of $\Gamma$ on $\mathbb C\setminus(-\mathbb N_0)$. Show that the function
$$\widetilde\Gamma(z)\widetilde\Gamma(1-z)-\frac{\pi}{\sin(\pi z)}$$
can be analytically continued to an entire function.

I do assume that the analytic continuation is the classical $$\widetilde\Gamma(z)=\frac{\Gamma(z+n)}{z(z+1)\cdot\ldots\cdot(z+n-1)}$$

for $z\in\mathbb C\setminus(-\mathbb N_0),\operatorname{Re} z>-n$ with residues $$\operatorname{Res}_{-n}(\widetilde\Gamma)=\frac{(-1)^n}{n!}$$ with $n\in\mathbb{N}_0$ which I had to deduce in the excersice leading to this problem. I stumbled upon explanations on how to compute $\widetilde\Gamma(z)\widetilde\Gamma(1-z)$ using the Beta function I am not familiar with. Now I am curious as to what I have to do exactly with the given function and which methods there are available to do so.

Answer

The "standard" analytic continuation leads to to Euler's limit product formula, that leads to the Weierstrass product:
$$ \Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n} \tag{1}$$
from which:
$$ z\,\Gamma(z)\Gamma(-z)=\frac{1}{z}\prod_{n\geq 1}\left(1-\frac{z^{2}}{n^2}\right)^{-1}\tag{2}$$
and you may finish the work by recognizing the (reciprocal) Weierstrass product for a sine function in the RHS.

As an alternative, you may just prove that

$\Gamma(z)\Gamma(1-z)-\frac{\pi}{\sin(\pi z)}$ has no singular point,
since both terms have simple poles at the same points with the same
residues: $$\text{Res}\left(\Gamma(z)\Gamma(1-z),z=-n\right)=(-1)^n,$$
$$\text{Res}\left(\frac{\pi}{\sin(\pi z)},z=-n\right)=\lim_{z\to
n}\frac{\pi(z-n)}{\sin(\pi z)}\stackrel{DH}{=}\cos(\pi
n)=(-1)^n.\tag{3}$$

Yet another way is to prove the red equality:
$$\frac{d^2}{dz^2}\log(\Gamma(z)\Gamma(1-z))=\psi''(z)+\psi''(1-z)=\sum_{n\geq 0}\left(\frac{1}{(z+n)^2}+\frac{1}{(1-z+n)^2}\right)\color{red}{=}\frac{\pi^2}{\sin^2(\pi z)}=\frac{d^2}{dz^2}\log\frac{\pi}{\sin(\pi z)}\tag{4}$$

through Fourier series or other means.