Weierstrass infinite product for the Gamma function

Weierstrass infinite product for the Gamma function

I want to derive in this post the Weierstrass infinite product gamma. It is a very useful expression to solve many problems involving the gamma function.

1.Weierstrass infinite product

1. The Weierstrass infinite product is given by

\frac{1}{\Gamma(z)}= z e^{\gamma z} \prod_{n=1}^{\infty}\Big(1+\frac{z}{n}\Big)e^{-\frac{z}{n}}

Recall Gauss infinite product for the Gamma function derived in this post 

\Gamma(z)= \lim_{n \rightarrow \infty} \,\frac{n!\, n^z}{z (z+1)\cdots (z+n)}(1.1)

Now,

\frac{1}{\Gamma(z)}=\lim_{n \rightarrow \infty} \,\frac{z (z+1)\cdots (z+n)}{n!\, n^z}

=\lim_{n \rightarrow \infty} \, \frac{1}{n^z}\cdot \frac{z}{n!} \cdot\Big( 1+\frac{z}{1}\Big)\cdot\ 2 \cdot \Big( 1+\frac{z}{2}\Big)\cdot 3 \cdot \Big( 1+\frac{z}{3}\Big) \cdots n\cdot\Big( 1+\frac{z}{n}\Big)

=\lim_{n \rightarrow \infty} \, \frac{z}{n^z}\cdot \frac{n!}{n!} \cdot\Big( 1+\frac{z}{1}\Big)\cdot\ \Big( 1+\frac{z}{2}\Big)\cdot \Big( 1+\frac{z}{3}\Big) \cdots \Big( 1+\frac{z}{n}\Big)

=\lim_{n \rightarrow \infty} \,z \cdot\frac{e^{z} \cdot e^{\frac{z}{2}} \cdot e^{\frac{z}{3}} \cdots e^{\frac{z}{n}} \cdot e^{-z} \cdot e^{\frac{z}{-2}} \cdot e^{\frac{-z}{3}} \cdots e^{\frac{-z}{n}} }{e^{z \log(n)}}\cdot\Big( 1+\frac{z}{1}\Big)\cdot\ \Big( 1+\frac{z}{2}\Big)\cdot \Big( 1+\frac{z}{3}\Big) \cdots \Big( 1+\frac{z}{n}\Big)

=\lim_{n \rightarrow \infty} \,z \cdot e^{z(1+ \frac{1}{2}+\frac{1}{3}+ \cdots + \frac{1}{n})} \cdot e^{-z \log(n)}\cdot\Big( 1+\frac{z}{1}\Big)\cdot\ \Big( 1+\frac{z}{2}\Big)\cdot \Big( 1+\frac{z}{3}\Big) \cdots \Big( 1+\frac{z}{n}\Big) \cdot e^{-z} \cdot e^{\frac{z}{-2}} \cdot e^{\frac{-z}{3}} \cdots e^{\frac{-z}{n}}

=\lim_{n \rightarrow \infty} \, e^{z(1+ \frac{1}{2}+\frac{1}{3}+ \cdots + \frac{1}{n}-\log(n))} \cdot z \cdot\Big( 1+\frac{z}{1}\Big) \cdot e^{-z} \cdot\ \Big( 1+\frac{z}{2}\Big) \cdot e^{\frac{z}{-2}}\cdot \Big( 1+\frac{z}{3}\Big) \cdot e^{\frac{-z}{3}} \cdots \Big( 1+\frac{z}{n}\Big) e^{\frac{-z}{n}}

=\lim_{n \rightarrow \infty} \, e^{z(1+ \frac{1}{2}+\frac{1}{3}+ \cdots + \frac{1}{n}-\log(n))} \cdot z \cdot \prod_{k=1}^{n}\Big( 1+\frac{z}{k}\Big) \cdot e^{- \frac{z}{k}}

and finally taking the limit!

\boxed{\frac{1}{\Gamma(z)}= z e^{\gamma z} \prod_{k=1}^{\infty}\Big(1+\frac{z}{k}\Big)e^{-\frac{z}{k}}}(1.2)

or

\Gamma(z)= \frac{e^{- \gamma z}}{z} \prod_{k=1}^{\infty}\Big(1+\frac{z}{k}\Big)^{-1}e^{\frac{z}{k}}

And recalling the functional equation  of the Gamma Function

\Gamma(z+1)=z \Gamma(z)

we get

\boxed{\Gamma(z+1)= \e^{- \gamma z} \prod_{k=1}^{\infty}\Big(1+\frac{z}{k}\Big)^{-1}e^{\frac{z}{k}}}(1.3)

2. Series Expansion Digamma function

If we take logs of (1.3)

\log(\Gamma(x+1))= -\gamma z - \sum_{k=1}^{\infty} \log \Big( 1+ \frac{1}{k}\Big)+\sum_{k=1}^{\infty} \frac{x}{k}

and differentiate with respect to x we get the Series representation of the digamma function, extremely useful

\boxed{\psi(x+1)=-\gamma+ \sum_{k=1}^{\infty} \Big( \frac{1}{k}-\frac{1}{k+x}\Big)}(2.1)

3. Series expansion Trigamma Function and the evaluation of an Infinite series

If we differentiate (2.1) we get

\boxed{\psi^{\prime}(x+1)=\sum_{k=1}^{\infty}\frac{1}{(k+x)^{2}}=\psi^{\prime}(x)-\frac{1}{x^2}}(3.1)

if we let x\rightarrow1-x and k\rightarrow-k-1 we get that

\psi^{\prime}(1-x)=\sum_{k=-\infty}^{-1}\frac{1}{(k+x)^2}

putting all together

\boxed{\psi^{\prime}(x)+\psi^{\prime}(1-x)=\sum_{k=-\infty}^{\infty}\frac{1}{(k+x)^2}}(3.2)

On the other hand we have the reflection formula for the Gamma function

\Gamma(x).\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}

taking log on both sides and differentiating w.r. to x we get

\psi(x)+\psi(1-x)=-\pi \cot( \pi x)

differentiating the expression above one more time

\boxed{\psi^{\prime}(x)+\psi^{\prime}(1-x)=\frac{\pi^2} {\sin^2( \pi x)}}(3.3)

Equating (3.2) and (3.3), we finally get

\boxed{\sum_{k=-\infty}^{\infty}\frac{1}{(k+x)^2}=\frac{\pi^2} {\sin^2( \pi x)}}

Ricardo Albahari

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