# Improper Integrals

• July 5th 2008, 11:33 AM
Aryth
Improper Integrals
There are several parts. Any help with any of them would be much appreciated:

a) By integration by parts show that

$\int^{\infty}_0 e^{-x^2}x^k~dx = \frac{k-1}{2}\int^{\infty}_0 e^{-x^2}x^{k-2}~dx$

b) Deduce from this recursion formula that:

$\int^{\infty}_0 e^{-x^2}x^{2m}~dx = \frac{1\cdot 3\cdot \cdot \cdot \cdot \cdot (2m-1)}{2^{m+1}}\sqrt{\pi}$

and

$\int^{\infty}_0 e^{-x^2}x^{2m-1}~dx = \frac{1}{2}(m-1)!$

c) Conclude that:

$\int^{\infty}_0 e^{-x^2}x^{n-1}~dx = \frac{\Gamma{(n/2)}}{2}$
• July 5th 2008, 11:57 AM
PaulRS
Quote:

Originally Posted by Aryth
There are several parts. Any help with any of them would be much appreciated:

a) By integration by parts show that

$\int^{\infty}_0 e^{-x^2}x^k~dx = \frac{k-1}{2}\int^{\infty}_0 e^{-x^2}x^{k-2}~dx$

b) Deduce from this recursion formula that:

$\int^{\infty}_0 e^{-x^2}x^{2m}~dx = \frac{1\cdot 3\cdot \cdot \cdot \cdot \cdot (2m-1)}{2^{m+1}}\sqrt{\pi}$

and

$\int^{\infty}_0 e^{-x^2}x^{2m-1}~dx = \frac{1}{2}(m-1)!$

c) Conclude that:

$\int^{\infty}_0 e^{-x^2}x^{n-1}~dx = \frac{\Gamma{(n/2)}}{2}$

a) note that: $
\int_0^\infty {e^{ - x^2 } \cdot x^{k} dx} =- \tfrac{1}
{2} \cdot \int_0^\infty {\left( {e^{ - x^2 } } \right)^\prime \cdot x^{k - 1} dx}
$

b) After proving a) proceed inductively remembering that $
\int_0^\infty {e^{ - x^2 } dx}
=\frac{\sqrt{\pi}}{2}$
and $
\int_0^\infty {e^{ - x^2 } \cdot x \cdot dx} = \tfrac{1}
{2}
$

Let's see: $
\tfrac{\pi }
{2} = \int_0^\infty {\tfrac{{dx}}
{{1 + x^2 }}} = \int_0^\infty {\int_0^\infty {e^{ - y \cdot \left( {1 + x^2 } \right)} dy} dx} = \int_0^\infty {e^{ - y} \left( {\int_0^\infty {e^{ - y \cdot x^2 } dx} } \right)dy}
$

Now: $
\int_0^\infty {e^{ - y \cdot x^2 } dx} = \frac{1}{\sqrt y} \cdot \int_0^\infty {e^{ - x^2 } dx}
$
so $
\int_0^\infty {e^{ - y} \left( {\int_0^\infty {e^{ - y \cdot x^2 } dx} } \right)dy} = \left( {\int_0^\infty {\tfrac{{e^{ - y} }}
{{\sqrt y }}dy} } \right)\left( {\int_0^\infty {e^{ - x^2 } dx} } \right)
$
but: $
\int_0^\infty {\tfrac{{e^{ - y} }}
{{\sqrt y }} \cdot dy} \underbrace = _{x = \sqrt y }2 \cdot \int_0^\infty {e^{ - x^2 } dx}
$
thus: $
\left( {\int_0^\infty {e^{ - x^2 } dx} } \right)^2 = \tfrac{\pi }
{4}
$