But why does any $\Gamma(z)\Gamma(1-z)\gets\pi\csc(\pi z)$?

Question

So I was looking through my questions for a sense of nostalgia when I came across this question of mine asking on how to evaluate $$\int_{-\infty}^\infty\Gamma(1+ix)\Gamma(1-ix)dx$$Now here's the thing: Here is a few steps that I took that I'm not really sure I understand:

$$\int_{-\infty}^\infty(ix)!(-ix)!dx$$$$\text{Which can be rewritten as }\int_{-\infty}^\infty ix\Gamma(ix)\Gamma(1-ix)dx$$$$\text{Which can then be simplified as }\int_{-\infty}^\infty ix\pi\csc(\pi ix)dx$$

Although here's the thing: why does any $\Gamma(z)\Gamma(1-z)$ simplify to $\pi\csc(\pi z)$? And when writing this question, I came across this question:

I'm reading on the extension of $\Gamma$ to the complex plane and there is written:

Corollary$$\Gamma(z)\ne0\qquad\forall z\in\mathbb{C}\setminus\{0,-1,-2,\dots\}$$

Proof

$(\forall z\in\mathbb{C}\setminus\mathbb{Z})\quad\Gamma(1-z)\Gamma(z)=\dfrac\pi{\sin(\pi z)}$ implies this. For $n\in\mathbb{N}$ is $\Gamma(n+1)=n!\ne0$

Why?

Can someone explain why this is true? Why does $\Gamma(1-z)\Gamma(z)=\dfrac\pi{\sin(\pi z)}$ imply $\Gamma(z)\ne0$?

As it turns out, it is because the RHS is a nonzero, so the LHS cannot be also. But that doesn't really answer my question.

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My question

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Why does $\Gamma(1-z)\Gamma(z)\gets\dfrac\pi{\sin(\pi z)}\implies\Gamma(z)\ne0$?

Answer 1

I think this is the Euler's Reflection formula and the proof begins with an integral representation of Beta functions;

$$B(a,b)=\frac{\Gamma_a \Gamma_b}{\Gamma_{a+b}}=\int_0^{\infty}\frac{x^{a-1}}{(1+x)^{a+b}}\, dx$$

Putting $a=z$ and $b=1-z$, we get

$${\Gamma_z \Gamma_{1-z}}=\int_0^{\infty}\frac{x^{z-1}}{(1+x)}\, dx$$

Splitting the integral and replacing $x$ with $\frac{1}{x}$ in the second integral;

$$=\int_0^{1}\frac{x^{z-1}}{(1+x)}\, dx + \int_1^{\infty}\frac{x^{z-1}}{(1+x)}\, dx$$

$$=\int_0^{1}\frac{x^{z-1}}{(1+x)}\, dx + \int_0^{1}\frac{x^{-z}}{(1+x)}\, dx$$

Using $\sum_{n=0}^{\infty} (-x)^n=\frac{1}{1+x} $ in the above integral;

$$=\int_0^{1}{x^{z-1}}\sum_{n=0}^{\infty} (-x)^n\, dx + \int_0^{1}{x^{-z}}\sum_{n=0}^{\infty} (-x)^n\, dx$$

$$=\sum_{n=0}^{\infty} (-1)^n \int_0^1 x^{n+z-1} \,dx + \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^{n-z} \,dx $$

$$=\sum_{n=0}^{\infty} (-1)^n \frac{1}{n+z} + \sum_{n=0}^{\infty} (-1)^n \frac{1}{n-z+1} $$

$$=\frac{1}{z}-\sum_{n=1}^{\infty} \frac{(-1)^n 2z}{n^2-z^2}.....(1)$$

$$=\frac{\pi}{\sin\pi z}$$

Edit: Regarding ($1$), there's another interesting proof referring to the fourier series of $\cos(zx)$, mentioned below;

$$\cos(zx)=\frac{2z\sin(\pi z)}{\pi} \left[\frac{1}{2z^2}-\sum_{n=1}^{\infty} \frac{(-1)^n \cos(nx)}{n^2-z^2}\right]$$