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Thread: 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

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

    b) Deduce from this recursion formula that:

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

    and

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

    c) Conclude that:

    $\displaystyle \int^{\infty}_0 e^{-x^2}x^{n-1}~dx = \frac{\Gamma{(n/2)}}{2}$
    Last edited by Aryth; Jul 5th 2008 at 11:38 AM. 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

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

    b) Deduce from this recursion formula that:

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

    and

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

    c) Conclude that:

    $\displaystyle \int^{\infty}_0 e^{-x^2}x^{n-1}~dx = \frac{\Gamma{(n/2)}}{2}$
    a) note that: $\displaystyle
    \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 $\displaystyle
    \int_0^\infty {e^{ - x^2 } dx}
    =\frac{\sqrt{\pi}}{2}$ and $\displaystyle
    \int_0^\infty {e^{ - x^2 } \cdot x \cdot dx} = \tfrac{1}
    {2}
    $

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