Where's the mistake in this proof of Euler's reflection formula?

Question

$$\frac{\sin(\pi x)}{\pi}=x\prod_{r=1}^\infty\left(1-\frac{x^2}{r^2}\right)$$ $$\Gamma(x)=\frac{1}{x}e^{-\gamma x}\prod_{r=1}^\infty\left(\frac{r}{x+r}\right)e^{\frac{x}{r}}$$ $$\Gamma(x) \Gamma(1-x)=\frac{1}{x(1-x)}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{r^2}{(x+r)(1-x+r)}\right)e^{\frac{1}{r}}$$ Assume $\Gamma(x) \Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$, then, for $x \notin \mathbb{Z}$:

$$\frac{1}{x}\prod_{r=1}^\infty\left(\frac{r^2}{r^2-x^2}\right)=\frac{1}{x(1-x)}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{r^2}{(x+r)(1-x+r)}\right)e^{\frac{1}{r}}$$ $$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{1}{1-x+r}\right)e^{\frac{1}{r}}$$ $$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty e^{\frac{1}{r}}\prod_{r=2}^\infty\left(\frac{1}{r-x}\right)$$ $$1=e^{-\gamma}\prod_{r=1}^\infty e^{\frac{1}{r}}$$ Thus $-\gamma+\sum_{r=1}^\infty \frac{1}{r}=0$, which is ridiculous. Where did I go wrong?

Answer 1

In the line

$$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{1}{1-x+r}\right)e^{\frac{1}{r}}$$

you have divergent products. If one wants to give meaning to it, the only somewhat sensible meaning is $0 = 0$. Of course dividing that equation by $0$ does not produce anything meaningful.

Answer 2

Instead of using the Weierstrass definiton of $\Gamma(1-z)$, rewrite $$\Gamma(1-z)=-z\Gamma(-z)$$ Rewriting it like that will short out out the $-z$ in the denominator and once you multiply it with $\Gamma(z)$ everything will cancel out leaving you with Euler's product definition of $\sin(\pi z)$ in the denominator and a $\pi$ in the numerator: $$\Gamma(z)\Gamma(1-z)=\Gamma(z)\Gamma(1+(-z))\\=-z\Gamma(z)\Gamma(-z)\\=\left(\frac{e^{-\gamma z}}{z}\prod_{k=1}^{\infty}(1+\frac zk)^{-1}e^{\frac zk}\right)\left(-z.\frac {e^{\gamma z}}{-z}\prod_{k=1}^{\infty}(1-\frac zk)^{-1}e^{\frac {-z}{k}}\right)=\frac{1}{z}\prod_{k=1}^{\infty}\left((1+\frac zk)(1- \frac zk)\right)^{-1}=\frac 1z\prod_{k=1}^{\infty}\left(1-\frac {z^{2}}{k^{2}}\right)^{-1}\\=\frac{\pi}{\pi z}\prod_{k=1}^{\infty}\left(1-\frac {(\pi z)^{2}}{\pi^{2} k^{2}}\right)^{-1}=\frac {\pi}{\sin(\pi z)}$$