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Math Help - Integral #5

  1. #1
    Super Member Random Variable's Avatar
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    Integral #5

    Challenge Problem:

     \int^{\infty}_{0} e^{-(x^{2}+ \frac{1}{x^{2}})} \ dx




    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; May 24th 2010 at 05:24 PM. Reason: Approval.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Challenge Problem:

     \int^{\infty}_{0} e^{-(x^{2}+ \frac{1}{x^{2}})} \ dx




    Moderator edit: Approved Challenge question.
    Let J=\int_0^\infty e^{x^2+\frac{1}{x^2}}dx split this into J=\int_0^1 e^{-(x^2+\frac{1}{x^2})}dx+\int_1^\infty e^{-(x^2+\frac{1}{x^2})}dx. In the first integral let x=\frac{1}{x} to get \int_0^1 \frac{1}{x^2}e^{-(x^2+\frac{1}{x^2})}dx and so J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x^2+\frac{1}{x^2})}dx. So, notice now that x+\frac{1}{x^2}=x^2-2x\frac{1}{x}+\frac{1}{x^2}+2=\left(x-\frac{1}{x}\right)^2+2. So, that J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x-\frac{1}{x})^2-2}. Let z=x-\frac{1}{x}\implies dz=\left(1+\frac{1}{x^2}\right)dx to get J=\int_0^\infty e^{-z^2-2}dz=\frac{\sqrt{\pi}}{2e^2}
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  3. #3
    Super Member Random Variable's Avatar
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    There always seems to be mutiple ways to solve the challenge problems I post. Here's what I did:


    Let  I(a) = \int^{\infty}_{0} e^{-(x^{2}+ \frac{a^{2}}{x^{2}})} \ dx

     I'(a) = - 2a \int^{\infty}_{0} \frac{1}{x^{2}} \ e^{-(x^{2}+  \frac{a^{2}}{x^{2}})} \ dx

    now let  u = \frac{a}{x}

     = - 2 \int^{\infty}_{0} e^{-(u^{2} + \frac{a^{2}}{x^{2}})} \ du = - 2 \  I(a)

    so  I(a) = Ce^{-2a}

    and since  I(0) =   C = \int^{\infty}_{0} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2}

     I(a) = \frac{\sqrt{\pi}}{2} e^{-2a}

    and  \int^{\infty}_{0} e^{-(x^{2}+ \frac{1}{x^{2}})} \ dx = I(1) =  \frac{\sqrt{\pi}}{2e^{2}}
    Last edited by Random Variable; May 24th 2010 at 09:20 PM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    There always seems to be mutiple ways to solve the challenge problems I post. Here's what I did:


    Let  I(a) = \int^{\infty}_{0} e^{-(x^{2}+ \frac{a^{2}}{x^{2}})} \ dx

     I'(a) = - 2a \int^{\infty}_{0} \frac{1}{x^{2}} \ e^{-(x^{2}+  \frac{a^{2}}{x^{2}})} \ dx

    now let  u = \frac{a}{x}

     = - 2 \int^{\infty}_{0} e^{-(u^{2} + \frac{a^{2}}{x^{2}})} \ du = - 2 \  I(a)

    so  I(a) = Ce^{-2x}

    and since  I(0) =   C = \int^{\infty}_{0} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2}

     I(a) = \frac{\sqrt{\pi}}{2} e^{-2x}

    and  \int^{\infty}_{0} e^{-(x^{2}+ \frac{1}{x^{2}})} \ dx = I(1) =  \frac{\sqrt{\pi}}{2e^{2}}
    Well, you bested me on this one! I was trying to look for some way to parameterize it but gave up. Good job!
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  5. #5
    Super Member Deadstar's Avatar
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    http://www.mathhelpforum.com/math-he...tml#post484278

    Similar question, used a different method.
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    Let J=\int_0^\infty e^{x^2+\frac{1}{x^2}}dx split this into J=\int_0^1 e^{-(x^2+\frac{1}{x^2})}dx+\int_1^\infty e^{-(x^2+\frac{1}{x^2})}dx. In the first integral let x=\frac{1}{x} to get \int_0^1 \frac{1}{x^2}e^{-(x^2+\frac{1}{x^2})}dx


    If you're making a variable change here (with the same name for both the original and the new variable...hmmm ), then the right hand integral must have limits between 1 and \infty , which is obvious from the continuation.
    Very nice trick.

    Tonio


    and so J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x^2+\frac{1}{x^2})}dx. So, notice now that x+\frac{1}{x^2}=x^2-2x\frac{1}{x}+\frac{1}{x^2}+2=\left(x-\frac{1}{x}\right)^2+2. So, that J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x-\frac{1}{x})^2-2}. Let z=x-\frac{1}{x}\implies dz=\left(1+\frac{1}{x^2}\right)dx to get J=\int_0^\infty e^{-z^2-2}dz=\frac{\sqrt{\pi}}{2e^2}
    .
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    .
    Yeah, sorry about the typo. It was late!
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  8. #8
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Deadstar View Post
    http://www.mathhelpforum.com/math-he...tml#post484278

    Similar question, used a different method.
    That's by far the easiest way.


    Proof that  \int^{\infty}_{-\infty} f \big(x - \frac{a}{x} \Big) \ dx  = \int^{\infty}_{-\infty} f(x)


     \int^{\infty}_{-\infty} f \big(x - \frac{a}{x} \Big) \ dx  = \lim_{b \to 0^{-}} \int^{b}_{-\infty} f \big(x - \frac{a}{x} \Big) \ dx + \lim_{ b \to 0^{+}} \int^{\infty}_{0} f \big(x - \frac{a}{x} \Big) \ dx


    let  u = x - \frac{a}{x}

     x^{2} - ux - a =0


    if  x<0, \ x = \frac{u - \sqrt{u^{2} + 4a}}{2}

    and  dx = \frac{1}{2} - \frac{u}{\sqrt{u^{2}+4a}}


    if  x >0, \ x=  \frac{u + \sqrt{u^{2} + 4a}}{2}

    and  dx = \frac{1}{2} + \frac{u}{\sqrt{u^{2}+4a}}


    so  \int^{\infty}_{-\infty} f \Big( x- \frac{a}{x} \Big) = \int^{\infty}_{-\infty} f(u) \Big( \frac{1}{2}- \frac{u}{\sqrt{u^{2}+4a}} \Big) \ du +\int^{\infty}_{-\infty} f(u) \Big( \frac{1}{2}+  \frac{u}{\sqrt{u^{2}+4a}} \Big) \ du  = \int^{\infty}_{-\infty} f(u) \ du
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