gamma function exercise asked in the math group by Chhaiyuth Chhel‎

Exercise:

Consider a Gamma function $\Gamma :{{\mathbb{R}}^{+}}\to \mathbb{R}$ defined to be

                           $\Gamma \left( x \right)=\int_{0}^{\infty }{{{t}^{x-1}}{{e}^{-t}}dt}\,\,\,,\,\,x>0$

1) Show that, $\Gamma \left( x \right)=\left( x-1 \right)\Gamma \left( x-1 \right)\,\,\forall x>1$ and $\Gamma \left( n \right)=\left( n-1 \right)!\,\,\,\,,\,\,\,n=1,2,...$

2) Show that, $\int_{0}^{\infty }{{{e}^{-{{x}^{2}}}}dx}=\frac{\sqrt{\pi }}{2}$ then deduce that $\Gamma \left( 1/2 \right)=\sqrt{\pi }$

Solution:

1) Let \(u={{t}^{x-1}}\,\,\And \,dv={{e}^{-t}}dt\Leftrightarrow du=\left( x-1 \right){{t}^{x-2}}dt\,\,\And \,\,v=-{{e}^{-t}}\)

So $\Gamma \left( x \right)=\left( -{{e}^{-t}}{{t}^{x-1}} \right)_{0}^{\infty }+\int_{0}^{\infty }{{{e}^{-t}}\left( x-1 \right){{t}^{x-2}}dt}$

But $\underset{t\to \infty }{\mathop{\lim }}\,{{t}^{x-1}}{{e}^{-t}}=0$ and $\underset{t\to 0}{\mathop{\lim }}\,{{e}^{-t}}{{t}^{x-1}}=0$

So $\Gamma \left( x \right)=\left( x-1 \right)\int_{0}^{\infty }{{{t}^{x-2}}{{e}^{-t}}dt}=\left( x-1 \right)\int_{0}^{\infty }{{{t}^{\left( x-1 \right)-1}}{{e}^{-t}}dt}=\left( x-1 \right)\Gamma \left( x-1 \right)$

Now if $x$is an integer $n=1,2,3,...$ then

$\Gamma \left( n \right)=\left( n-1 \right)\Gamma \left( n-1 \right)=\left( n-1 \right)\left( n-2 \right)\Gamma \left( n-2 \right)=\left( n-1 \right)\left( n-2 \right)...1=\left( n-1 \right)!$

We will proof this result using induction as follows:

For $n=1\Leftrightarrow \Gamma \left( 1 \right)=\left( 1-1 \right)!=0!=1$

For $n=2\Leftrightarrow \Gamma \left( 2 \right)=\left( 2-1 \right)\Gamma \left( 2-1 \right)=1\Gamma \left( 1 \right)=1=1!$

For $n=3\Leftrightarrow \Gamma \left( 3 \right)=\left( 3-1 \right)\Gamma \left( 3-1 \right)=2\Gamma \left( 2 \right)=2=2!$

Suppose the statement is true for $n=k$ , $\Gamma \left( k \right)=\left( k-1 \right)!$ and lets prove the statement is also true for $n=k+1$

$\Gamma \left( k+1 \right)=\left( k+1-1 \right)\Gamma \left( k+1-1 \right)=k\Gamma \left( k \right)=k\left( k-1 \right)\times \left( k-1 \right)\times ...\times 2\times 1=k!$

2) Let $I=\int_{0}^{\infty }{{{e}^{-{{x}^{2}}}}dx}\Leftrightarrow {{I}^{2}}=\left( \int_{0}^{\infty }{{{e}^{-{{x}^{2}}}}dx} \right)\left( \int_{0}^{\infty }{{{e}^{-{{y}^{2}}}}}dy \right)$

So ${{I}^{2}}=\int_{0}^{\infty }{\int_{0}^{\infty }{{{e}^{-\left( {{x}^{2}}+{{y}^{2}} \right)}}dxdy}}$ using polar coordinates to get

${{I}^{2}}=\int_{0}^{\frac{\pi }{2}}{\left( \int_{0}^{\infty }{{{e}^{-{{r}^{2}}}}}rdr \right)d\theta }$

Let $u={{r}^{2}}\Leftrightarrow du=2rdr\Leftrightarrow \frac{du}{2}=rdr$

So $\int_{0}^{\infty }{{{e}^{-{{r}^{2}}}}rdr}=\frac{1}{2}\int_{0}^{\infty }{{{e}^{-{{r}^{2}}}}d\left( {{r}^{2}} \right)}=-\frac{1}{2}\left( {{e}^{-{{r}^{2}}}} \right)_{0}^{\infty }=-\frac{1}{2}\left( {{e}^{-\infty }}-{{e}^{0}} \right)=\frac{1}{2}$

Hence ${{I}^{2}}=\int_{0}^{\frac{\pi }{2}}{\int_{0}^{\infty }{{{e}^{-{{r}^{2}}}}rdrd\theta }}=\int_{0}^{\frac{\pi }{2}}{\frac{1}{2}d\theta =\frac{\pi }{4}}$ thus $I=\frac{\sqrt{\pi }}{2}$

\[\Gamma \left( \frac{1}{2} \right)=\int_{0}^{\infty }{{{t}^{1/2-1}}{{e}^{-t}}dt}=\int_{0}^{\infty }{{{t}^{-1/2}}{{e}^{-t}}dt}=\int_{0}^{\infty }{\frac{{{e}^{-t}}}{\sqrt{t}}dt}\]

Let $u=\sqrt{t}\Leftrightarrow du=\frac{1}{2\sqrt{t}}dt\Leftrightarrow 2udu=dt$

So $\int_{0}^{\infty }{\frac{{{e}^{-t}}}{\sqrt{t}}dt}=\int_{0}^{\infty }{2u\frac{{{e}^{-{{u}^{2}}}}}{u}du}=2\int_{0}^{\infty }{{{e}^{-{{u}^{2}}}}du}=2\left( \frac{\sqrt{\pi }}{2} \right)=\sqrt{\pi }$