Abstract Nonsense

Crushing one theorem at a time

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

\text{ }

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

\text{ }

where \gamma is the Euler-Maschernoi constant.

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

From this the rest is just computation:

\text{ }

\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}

\text{ }

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))}

\text{ }

converges to a non-zero value and thus

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

converges. So, taking n large enough so that1+\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

\text{ }

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

\text{ }

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

\text{ }

\text{ }

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.


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December 28, 2010 - Posted by | Analysis, Computations, Fun Problems, Uncategorized | , ,

2 Comments »

  1. […] do this we first recall that for we […]

    Pingback by An Interesting Sum Involving Riemann’s Zeta Function « Abstract Nonsense | December 28, 2010 | Reply

  2. […] prove the second equality in (5) refer to this post from the Abstract Nonsense blog. The third equality follows similarly from the definition of the […]

    Pingback by A Collection of Infinite Products – I « Chaitanya's Random Pages | August 5, 2011 | Reply


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