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Math Help - Solving integrals using Gamma functions

  1. #1
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    Solving integrals using Gamma functions

    <br />
 \int_{0}^{\infty} \cos (x^{2}) dx = \int_{0}^{\infty} \sin (x^{2}) dx = \frac{1}{2}\sqrt{\frac{\pi}{2}} <br />

    Can this be proved using Gamma functions without using complex numbers or De Moivre's Theorem? I'm doing problems on Gamma and Beta functions yet and I have this problem in my exercise and I haven't covered extension of integrals to complex numbers yet so I'm assuming that you can do this without using complex numbers but I fail to understand how!
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  2. #2
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    Perhaps I am driving on the detour , you will see that I am actually making use of the calculus of the parameter , together with Laplace Transform .


    Let's consider  \int_0^{\infty} \sin(x^2) ~dx , first we make a substitution  x = \sqrt{t} then we have

     \frac{1}{2} ~ \int_0^{\infty}  \frac{\sin(t)}{\sqrt{t}}~dt


    at this moment i remember that the Laplace Transform of  f(x) =  \frac{1}{\sqrt{x}} is  \frac{ \sqrt{\pi} }{\sqrt{s} } by changing the dummy variables we have :


     \frac{1}{\sqrt{t} } = \frac{1}{\sqrt{\pi}} ~ \int_0^{\infty} e^{-ty} \frac{dy}{\sqrt{y} }

    Following it is the change of the order of integration :

      \frac{1}{2\sqrt{\pi}} ~  \int_0^{\infty} \frac{1}{\sqrt{y}} \int_0^{\infty} e^{-ty} \sin(t) ~dt~dy


     = \frac{1}{2\sqrt{\pi}} ~  \int_0^{\infty} \frac{1}{y^2 + 1 } ~\frac{dy}{\sqrt{y}}

    Sub.  y = u^2

     = \frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{du}{u^4  + 1 }


    Consider  I =  \int_0^{\infty} \frac{u^2 - 1 }{u^4  +  1} ~du by changing  u = 1/t we have  I = - I so  I = 0 . Also , the similar integral but which is not zero again :

     \int_0^{\infty} \frac{u^2 + 1 }{u^4  +  1} ~du = \int_0^{\infty} \frac{1+ 1/u^2}{ (u-1/u)^2 + 2 }~du  , sub  u-1/u = x

     = \int_{-\infty}^{\infty} \frac{dx}{x^2  + 2 } = \frac{\pi}{\sqrt{2}}

    By considering the difference of two integrals we have

     2 \int_0^{\infty} \frac{du}{u^4  + 1 } = \frac{\pi}{\sqrt{2}}

    or

     \int_0^{\infty} \frac{du}{u^4  + 1 } = \frac{\pi}{2\sqrt{2}}


    Finally ,   \int_0^{\infty} \sin(x^2) ~dx  =  \frac{\pi}{2\sqrt{\pi}~\sqrt{2}}  =  \frac{\sqrt{\pi}}{2\sqrt{2}}



    Remarks :

    If we consider the sum of these two integrals , we obtain


     \int_0^{\infty} \frac{u^2}{u^4  + 1 } ~du  = \frac{\pi}{2\sqrt{2}}  this result helps evaluate the integral  \int_0^{\infty} \cos(x^2)~dx which can be started in a quite similar way .
    Last edited by simplependulum; September 12th 2010 at 03:08 AM.
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  3. #3
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    Thanks! I hadn't thought of Laplace transforms at all.
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