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Math Help - Improper Integrals

  1. #1
    Super Member Aryth's Avatar
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    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}
    Last edited by Aryth; July 5th 2008 at 12:38 PM. Reason: I edited it. Everything should be correct now.
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  2. #2
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Aryth View Post
    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: <br />
\int_0^\infty  {e^{ - x^2 }  \cdot x^{k} dx}  =- \tfrac{1}<br />
{2} \cdot \int_0^\infty  {\left( {e^{ - x^2 } } \right)^\prime   \cdot x^{k - 1} dx} <br />

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

    Let's see: <br />
\tfrac{\pi }<br />
{2} = \int_0^\infty  {\tfrac{{dx}}<br />
{{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} <br />

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