Deriving $\sin(\pi s)=\pi s\prod_{n=1}^\infty (1-\frac{s^2}{n^2})$ without Hadamard Factorization I want to prove the Euler Product Formula for the sine function without Hadamard Factorization Theorem.
I know that $$\Gamma(s)\Gamma(1-s)=\prod_{n=1}^\infty (1-\frac{s^2}{n^2})^{-1}$$
And $$\Gamma(\frac{2n+1}{2})\Gamma(1-\frac{2n+1}{2})=(-1)^n\pi$$
So if $$\Gamma(s)\Gamma(1-s)=\frac{\pi}{f(s)}$$
$f(s)$ has a first order zero at every integer,  $f(\frac{2n+1}{2})=(-1)^n$, and $|f(s)|\le 1$.
Question: Is there a way to prove that $f(s)=\sin(\pi s)$
 is the only function to satisfy those conditions?
 A: In fact, it is possible to determine that denominator by evaluating $\Gamma(s)\Gamma(1-s)$ using the Gamma's integral definition:
$$
\Gamma(s)\triangleq\int_0^\infty x^{s-1}e^{-x}\mathrm dx
$$
Without loss of generality, we may set $\Re(s)\in(0,1)$:
$$
\begin{aligned} \Gamma(s)\Gamma(1-s) &=\int_0^\infty x^{s-1}e^{-x}\mathrm dx\int_0^\infty y^{-s}e^{-y}\mathrm dy \\ &=\int_0^\infty\int_0^\infty(x/y)^se^{-(x+y)}x^{-1}\mathrm dx\mathrm dy \end{aligned}
$$
Now, let $u=x/y,v=x+y$, then the Jacobian determinant is ${v\over(u+1)^2}$; as a result, the integral becomes
$$
\begin{aligned} \Gamma(s)\Gamma(1-s) &=\int_0^\infty\int_0^\infty u^se^{-v}\cdot{u+1\over uv}\cdot{v\over(u+1)^2}\mathrm du\mathrm dv \\ &=\int_0^\infty\int_0^\infty u^{s-1} e^{-v}\cdot{1\over u+1}\mathrm du\mathrm dv \\ &=\int_0^\infty e^{-v}\mathrm dv\int_0^\infty{u^{s-1}\over u+1}\mathrm du \\ &=\int_0^\infty{u^{s-1}\over u+1}\mathrm du \end{aligned}
$$
To evaluate this improper integral, we use contour integration:
$$
\oint_C{z^{s-1}\over z+1}\mathrm dz=\int_\color{blue}{\gamma_1}{z^{s-1}\over z+1}\mathrm dz+\int_\Gamma{z^{s-1}\over z+1}\mathrm dz+\int_\color{orange}{\gamma_2}{z^{s-1}\over z+1}\mathrm dz+\int_\color{red}\Sigma{z^{s-1}\over z+1}\mathrm dz \tag3
$$
where the contour is shown below:

For the right hand side, plugging in the parametrization for the line integrals gives
$$
\color{blue}{\gamma_1}:z=t,\mathrm dz=\mathrm dt,t:\epsilon e^{i\alpha}\to Re^{i\beta} \\ \int_\color{blue}{\gamma_1}{z^{s-1}\over z+1}\mathrm dz=\int_{\epsilon e^{i\alpha}}^{Re^{i\beta}}{t^{s-1}\over t+1}\mathrm dt
$$
$$
\Gamma:z=Re^{i\theta},\mathrm dz=iRe^{i\theta}\mathrm d\theta,\theta:\beta\to2\pi-\beta \\ \begin{aligned} \int_\Gamma{z^{s-1}\over z+1}\mathrm dz &=iR^s\int_{\beta}^{2\pi-\beta}{e^{i\theta s}\over Re^{i\theta}+1}\mathrm d\theta \\ &=i(2\pi-2\beta){R^se^{i\xi s}\over Re^{i\xi}+1} \end{aligned} \\
$$
$$
\color{orange}{\gamma_2}:z=t,\mathrm dz=\mathrm dt,t:Re^{i(2\pi-\beta)}\to\epsilon e^{i(2\pi-\alpha)} \\ \int_\color{orange}{\gamma_2}{z^{s-1}\over z+1}\mathrm dz=\int_{Re^{i(2\pi-\beta)}}^{\epsilon e^{i(2\pi-\alpha)}}{t^{s-1}\over t+1}\mathrm dt
$$
$$
\color{red}\Sigma:z=\epsilon e^{i\theta},\mathrm dz=i\epsilon e^{i\theta}\mathrm d\theta,\theta:2\pi-\alpha\to\alpha \\ \begin{aligned} \int_\color{red}\Sigma{z^{s-1}\over z+1}\mathrm dz &=\epsilon^s\int_{2\pi-\alpha}^\alpha{e^{i\theta s}\over \epsilon e^{i\theta}+1}\mathrm d\theta \\ &=(2\pi-2\alpha){\epsilon^se^{i\eta s}\over\epsilon e^{i\eta}+1} \end{aligned}
$$
For the left hand side, applying residue theorem yields
$$
\oint_C{z^{s-1}\over z+1}\mathrm dz=2\pi i\lim_{z\to-1}z^{s-1}=2\pi ie^{i\pi(s-1)}=-2\pi e^{i\pi s}
$$
Putting them altogether gives
$$
-2\pi ie^{i\pi s}=\int_{\epsilon e^{i\alpha}}^{Re^{i\beta}}{t^{s-1}\over t+1}\mathrm dt+i(2\pi-2\beta){R^se^{i\xi s}\over Re^{i\xi}+1}+\underbrace{\int_{Re^{i(2\pi-\beta)}}^{\epsilon e^{i(2\pi-\alpha)}}{t^{s-1}\over t+1}\mathrm dt}_{t\mapsto e^{2\pi i}x}+(2\pi-2\alpha){\epsilon^se^{i\eta s}\over\epsilon e^{i\eta}+1}
$$
After some algebraic manipulation, we have
$$
-2\pi ie^{i\pi s}=(1-e^{2\pi is})\int_0^\infty{t^{s-1}\over t+1}\mathrm dt \\ \therefore\int_0^\infty{t^{s-1}\over t+1}\mathrm dt={2\pi ie^{i\pi s}\over e^{2\pi is}-1}={2\pi i\over e^{\pi is}-e^{-\pi is}}={\pi\over\sin(\pi s)}
$$
which completes the derivation.
