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Given $g(x)=\frac{x}{x!(-x)!}$ equals $0$ for all integer values of $x$ and the functions $\sin(\pi x)$ and $\cos(\frac{\pi}{2x})$ having the same set of roots. Also knowing $(\frac{1}{2})!=\frac{\sqrt\pi}{2}$, derive a striking relationship from the previous conditions between the factorial function a trigonometric function.

I read a proof for this which derived $$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$$ but the proof was rather unsatisfying and failed to use the aforementioned conditions. My derivations has remained unsuccessful, can someone please provide a clever proof? Also, list the thought process of your approach to the question!

thanks in advance

Blue
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Philip Shen
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1 Answers1

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Note that the original definition Euler came up with for $\Gamma(x)$ before the integral form was actually this, $$\lim_{N\to\infty}N^{x-1}\prod_{k=1}^N\frac{k}{k+x-1}$$ which I previously derived in this post: Derivatives or Integrals of Factorials

You can just plug in this definition in $$\frac{1}{\Gamma(x)\Gamma(1-x)}$$ and after some simplification you get the following product, $$x\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2}\right)$$ Euler also showed that, $$\sin x=x\prod_{k=1}^{\infty}\left(1-\frac{x^2}{\pi^2k^2}\right)$$ Plugging in $\pi x$ for $x$ in the above formula (called the "sine product formula" which you can easily find a proof of, on google) gives us, $$\frac{\sin\pi x}{\pi}=x\prod_{k=1}^{\infty}\left(1-\frac{\pi^2x^2}{\pi^2k^2}\right)$$ $$\frac{\sin\pi x}{\pi}=x\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2}\right)$$ Which happens to be exactly the product we got after simplifying $$\frac{1}{\Gamma(x)\Gamma(1-x)}$$ This implies that, $$\frac{\sin\pi x}{\pi}=\frac{1}{\Gamma(x)\Gamma(1-x)}$$ $$\frac{\pi}{\sin\pi x}=\Gamma(x)\Gamma(1-x)$$

wolfqz
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