• Nov 6th 2009, 01:35 AM
chisigma
Given ...

$\displaystyle \gamma(z)= \int_{0}^{\infty} t^{z}\cdot e^{-t}\cdot dt$ (1)

... and the Euler's constant...

$\displaystyle \gamma = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n} \frac{1}{k} - \ln n \}= .577215664901...$ (2)

... demonstrate from definition (2) that is...

$\displaystyle \frac{1}{\gamma(z)} = e^{\gamma z}\cdot \prod_{n=1}^{\infty} (1+ \frac{z}{n})\cdot e^{-\frac{z}{n}}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 6th 2009, 08:05 AM
Drexel28
Quote:

Originally Posted by chisigma
Given ...

$\displaystyle \gamma(z)= \int_{0}^{\infty} t^{z}\cdot e^{-t}\cdot dt$ (1)

... and the Euler's constant...

$\displaystyle \gamma = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n} \frac{1}{k} - \ln n \}= .577215664901...$ (2)

... demonstrate from definition (2) that is...

$\displaystyle \frac{1}{\gamma(z)} = e^{\gamma z}\cdot \prod_{n=1}^{\infty} (1+ \frac{z}{n})\cdot e^{-\frac{z}{n}}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Problem: Prove that if $\displaystyle \gamma(z)=\int_0^{\infty}t^ze^{-t}dt=\Gamma(z+1)$ then find $\displaystyle \frac{1}{\gamma(z)}$

Proof:

Lemma: Let $\displaystyle \Gamma_\sigma=\frac{\sigma!\sigma^x}{x\cdot(1+x)\c dots(\sigma+x)}$ then $\displaystyle \Gamma(x)=\lim_{\sigma\to\infty}\Gamma_{\sigma}(x)$

Proof: It can be readily seen that $\displaystyle \Gamma_{\sigma}(x+1)=\frac{\sigma!\sigma^{x+1}}{(x +1)(x+2)\cdots(x+r+1)}=\frac{\sigma\cdot x}{\sigma+x+1}\Gamma_\sigma(x)$. So that $\displaystyle \Gamma(x+1)=\lim_{\sigma\to\infty}\Gamma_{\sigma}( x+1)=\lim_{\sigma\to\infty}\left\{\frac{\sigma\cdo t x}{\sigma+x+1}\Gamma_{\sigma}(x)\right\}=x\Gamma(x )$. Furthermore $\displaystyle \lim_{\sigma\to\infty}\Gamma_{\sigma}(1)=\lim_{\si gma\to\infty}\frac{\sigma}{\sigma+1}=1$. Thus the functional equation and the boundary condition are satisfied $\displaystyle \quad\blacksquare$

Now using this we can get somehwere. If we look closely at $\displaystyle \Gamma_{\sigma}(x)$ we can see that $\displaystyle \Gamma_{\sigma}(x)=\frac{\sigma!\sigma^{x}}{(x\cdo t(1+x)\cdots(\sigma+x)}=\frac{\sigma^x}{\frac{1}{\ sigma!}x\cdot(1+x)\cdots(\sigma+x)}$$\displaystyle =\frac{\sigma^x}{\frac{1}{1\cdots\sigma}x\cdot(1+x )\cdots(\sigma+x)}=\frac{\sigma^x}{x\cdot\left(1+\ frac{x}{1}\right)\cdots\left(1+\frac{x}{\sigma}\ri ght)}. Now using the common fact that \displaystyle \sigma^x=e^{x\ln(\sigma)} yields \displaystyle \Gamma_{\sigma}(x)=\frac{e^{x\ln(\sigma)}}{x\cdot\ left(1+\frac{x}{1}\right)\cdots\left(1+\frac{x}{\s igma}\right)}. Now using a litle snake-oil we can see that \displaystyle x\cdot\ln(\sigma)=x\left(\ln(\sigma)-H_{\sigma}+H_{\sigma} \right)=x\left(\ln(\sigma)-H_{\sigma}\right)+x\cdot H_{\sigma} where \displaystyle H_\sigma=\sum_{\ell=1}^{\sigma}\frac{1}{\ell}. So then \displaystyle \Gamma_{\sigma}(x)=\frac{e^{x\left(\ln(\sigma)-H_{\sigma}\right)}e^{x\cdot H_\sigma}}{x\cdot\left(1+\frac{x}{1}\right)\cdots\ left(1+\frac{x}{\sigma}\right)} or equivalently \displaystyle \Gamma_{\sigma}(x)=\frac{e^{-x\left(H_\sigma-\ln(\sigma)\right)}}{x}\cdot\prod_{\ell=1}^{\sigma }\left\{\frac{e^{\frac{x}{\ell}}}{1+\frac{x}{\ell} }\right\}. So that \displaystyle \frac{1}{\Gamma(x)}=\lim_{\sigma\to\infty}\frac{1} {\Gamma_{\sigma}(x)}=\lim_{\sigma\to\infty}\left\{ xe^{x\left(H_{\sigma}-\ln(\sigma)\right)}\prod_{\ell=1}^{\sigma}\left\{\ left(1+\frac{x}{\ell}\right)e^{\frac{-x}{\ell}}\right\}\right\}$$\displaystyle =xe^{\gamma x}\prod_{\ell=1}^{\infty}\left\{\left(1+\frac{x}{\ ell}\right)e^{\frac{-x}{\ell}}\right\}\quad\blacksquare$
• Nov 9th 2009, 12:01 AM
chisigma
The 'lemma' ...

$\displaystyle \gamma(z)= \lim_{k \rightarrow \infty} \gamma_{k} (z)$ (1)

... where...

$\displaystyle \gamma_{k} (z)= \frac{(k+1)!\cdot k^{z}}{(z+1)\cdot (z+2)\dots (z+k)\cdot (z+k+1)}$ (2)

... must be demonstrated and that can be done with the 'little nice formula' discussed in a previous thread...

$\displaystyle \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} k\cdot \int_{0}^{1} f(ku)\cdot (1-u)^{k}\cdot du$ (3)

If we compute the integral (3) with $\displaystyle f(t)=t^{z}$ applying systematically the integration by parts we obtain...

$\displaystyle \int_{0}^{\infty} t^{z}\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} k^{1+z}\cdot \int_{0}^{1} ku^{z}\cdot (1-u)^{k}\cdot du =$

$\displaystyle \lim_{k \rightarrow \infty} \frac{(k+1)!\cdot k^{z}}{(z+1)\cdot (z+2)\dots (z+k)\cdot (z+k+1)}=$

$\displaystyle \lim_{k \rightarrow \infty} \frac{k^{z}}{(1+z)\cdot (1+\frac{z}{2})\dots (1+\frac{z}{k})\cdot (1+\frac{z}{k+1})}$ (4)

Now if we take into account the identity...

$\displaystyle k^{z}= e^{z \ln k}= e^{z(\ln k -1-\frac{1}{2} -\dots -\frac{1}{k})}\cdot e^{z(1+\frac{1}{2} +\dots +\frac{1}{k})}$ (5)

... from (4) we derive the 'infinite product'...

$\displaystyle \frac{1}{\gamma (z)}= e^{\gamma z}\cdot \prod_{k=1}^{\infty} (1+\frac{z}{k})\cdot e^{-\frac{z}{k}}$ (6)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 9th 2009, 03:30 AM
Drexel28
Quote:

Originally Posted by chisigma
The 'lemma' ...

$\displaystyle \gamma(z)= \lim_{k \rightarrow \infty} \gamma_{k} (z)$ (1)

... where...

$\displaystyle \gamma_{k} (z)= \frac{(k+1)!\cdot k^{z}}{(z+1)\cdot (z+2)\dots (z+k)\cdot (z+k+1)}$ (2)

... must be demonstrated and that can be done with the 'little nice formula' discussed in a previous thread...

$\displaystyle \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} k\cdot \int_{0}^{1} f(ku)\cdot (1-u)^{k}\cdot du$ (3)

If we compute the integral (3) with $\displaystyle f(t)=t^{z}$ applying systematically the integration by parts we obtain...

$\displaystyle \int_{0}^{\infty} t^{z}\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} k^{1+z}\cdot \int_{0}^{1} ku^{z}\cdot (1-u)^{k}\cdot du =$

$\displaystyle \lim_{k \rightarrow \infty} \frac{(k+1)!\cdot k^{z}}{(z+1)\cdot (z+2)\dots (z+k)\cdot (z+k+1)}=$

$\displaystyle \lim_{k \rightarrow \infty} \frac{k^{z}}{(1+z)\cdot (1+\frac{z}{2})\dots (1+\frac{z}{k})\cdot (1+\frac{z}{k+1})}$ (4)

Now if we take into account the identity...

$\displaystyle k^{z}= e^{z \ln k}= e^{z(\ln k -1-\frac{1}{2} -\dots -\frac{1}{k})}\cdot e^{z(1+\frac{1}{2} +\dots +\frac{1}{k})}$ (5)

... from (4) we derive the 'infinite product'...

$\displaystyle \frac{1}{\gamma (z)}= e^{\gamma z}\cdot \prod_{k=1}^{\infty} (1+\frac{z}{k})\cdot e^{-\frac{z}{k}}$ (6)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Why must it be 'demonstrated'? I in fact showed that it is equivalent.