# Abstract Nonsense

## Infinite Product Representation of the Gamma Function

Point of post: In this post we prove that for every $x\in\mathbb{R}-\left(-\mathbb{N}\right)$ we have that

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

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where $\gamma$ is the Euler-Maschernoi constant.

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Theorem: Let $x\in\mathbb{R}-\left(-\mathbb{N}\right)$. Then,

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

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Proof: We recall the well-known Euler definition of the Gamma function, namely for $x\in\mathbb{R}-\left(-\mathbb{N}\right)$ we have that

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$\displaystyle \Gamma(x)=\lim_{k\to\infty}\frac{k^x}{x(1+x)(1+\frac{x}{2})\cdots(1+\frac{x}{k})}$

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From this the rest is just computation:

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\displaystyle \begin{aligned}\Gamma(x) &= \lim_{k\to\infty}\frac{k^x}{x(1+x)(1+\frac{x}{2})\cdots(1+\frac{x}{k})}\\ &= \lim_{k\to\infty}\frac{k^x}{x}\prod_{n=1}^{k}\left(1+\frac{x}{n}\right)^{-1}\\ &= \lim_{k\to\infty}\frac{e^{x\log(k)}}{x}\prod_{n=1}^{k}\left(1+\frac{x}{n}\right)^{-1}\\ &= \lim_{k\to\infty}\frac{e^{x\left(\log(k)-H_k+H_k\right)}}{x}\prod_{n=1}^{k}\left(1+\frac{x}{n}\right)^{-1}\\ &= \lim_{k\to\infty}\frac{e^{-x(H_k-\log(k))}}{x}\prod_{n=1}^{k}\left(1+\frac{x}{n}\right)^{-1}e^{\frac{x}{n}}\\ &= \lim_{k\to\infty}\frac{e^{-x(H_k-\log(k))}}{x}\prod_{k=1}^{\infty}\left(1+\frac{x}{n}\right)^{-1}e^{\frac{x}{n}}\\ &= \frac{e^{-\gamma x}}{x}\prod_{n=1}^{\infty}\left(1+\frac{x}{n}\right)^{-1}e^{\frac{x}{n}}\end{aligned}

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where the only part lacking possible rigor is the splitting of the limit over the product. But, this is justified since trivially

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$\displaystyle \lim_{k\to\infty}e^{-x(H_k-\log(k))}$

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converges to a non-zero value and thus

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$\displaystyle \lim_{k\to\infty}\prod_{n=1}^{k}\left(1+\frac{x}{n}\right)^{-1}e^{\frac{x}{n}}$

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must converge  since the whole limit converges. Or, if this doesn’t quite meet your standards recall that for a sequence of positive constants $a_n$ one has

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$\displaystyle \prod_{n\in\mathbb{N}}a_n\text{ converges if and only if }\sum_{n\in\mathbb{N}}\ln(a_n)$

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converges. So, taking $n$ large enough so that$1+\frac{x}{n}>0$ we may apply this and note that

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$\displaystyle -\log\left(1+\frac{x}{n}\right)-\frac{x}{n}\sim\frac{x}{n^2}$

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either way, we’re done. $\blacksquare$

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Corollary: Let $x\in\mathbb{R}-\mathbb{N}$, then

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$\displaystyle \Gamma(-x)=\frac{e^{\gamma x}}{-x}\prod_{n=1}^{\infty}\left(1-\frac{x}{n}\right)^{-1}e^{\frac{-x}{n}}$

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References:

1. Apostol, Tom M. Mathematical Analysis. London: Addison Wesley Longman, 1974. Print.

2. Edwards, Harold M. Riemann’s Zeta Function. Mineola, NY: Dover Publications, 2001. Print.