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.
 A: 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.
A: The Laplace transform of $t^{-1+\alpha}$ for $\alpha > 0$ is given by the following for $s > 0$:
\begin{align}
  \mathscr{L}\{t^{-1+\alpha}\} & = \int_{0}^{\infty}e^{-st}t^{-1+\alpha}dt \\
    & = \int_{0}^{\infty}e^{-st}(st)^{-1+\alpha}d(st)\cdot s^{-\alpha} \\
    & = \int_{0}^{\infty}e^{-u}u^{-1+\alpha}du\cdot s^{-\alpha}=\Gamma(\alpha)s^{-\alpha}
\end{align}
Because of this,
\begin{align}
 \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\mathscr{L}\{ t^{-1+\alpha+\beta} \} 
 & = \Gamma(\alpha)\Gamma(\beta)s^{-\alpha-\beta} \\
   & = \Gamma(\alpha)s^{-\alpha}\Gamma(\beta)s^{-\beta} \\
   & = \mathscr{L}\{ t^{-1+\alpha}\}\mathscr{L}\{ t^{-1+\beta}\} \\
   & = \mathscr{L}\{ t^{-1+\alpha} \star t^{-1+\beta} \}
\end{align}
By uniqueness of transforms, the convolution of $t^{-1+\alpha}$ and $t^{-1+\beta}$ is given by
$$
     t^{-1+\alpha}\star t^{-1+\beta} = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}t^{-1+\alpha+\beta},\;\;\; \alpha,\beta > 0.
$$
If $0 < \alpha < 1$ and $\beta = 1-\alpha$, then
$$
       t^{-1+\alpha}\star t^{-\alpha} = \Gamma(\alpha)\Gamma(1-\alpha)
$$
Curiously, the above does not depend on $t > 0$. So, set $t=1$:
$$
     \Gamma(\alpha)\Gamma(1-\alpha)=\int_{0}^{1}u^{-1+\alpha}(1-u)^{-\alpha}du.
$$
By turning the integral on the right into a contour integral enclosing $[0,1]$ in its interior, you can show that $F(\alpha)=\sin(\pi\alpha)\Gamma(\alpha)\Gamma(1-\alpha)$ is a contour integral that has an entire extension. That's enough for what you want. In fact, you can then evaluate the contour integral by a single residue at $\infty$ to obtain the full identity
$$
           \Gamma(z)\Gamma(1-z)\sin(\pi z) = \pi.
$$
