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Thread: Solving integrals using Gamma functions

  1. #1
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    Solving integrals using Gamma functions

    $\displaystyle
    \int_{0}^{\infty} \cos (x^{2}) dx = \int_{0}^{\infty} \sin (x^{2}) dx = \frac{1}{2}\sqrt{\frac{\pi}{2}}
    $

    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 $\displaystyle \int_0^{\infty} \sin(x^2) ~dx $ , first we make a substitution $\displaystyle x = \sqrt{t} $ then we have

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


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


    $\displaystyle \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 :

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


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

    Sub. $\displaystyle y = u^2 $

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


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

    $\displaystyle \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 $\displaystyle u-1/u = x $

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

    By considering the difference of two integrals we have

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

    or

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


    Finally , $\displaystyle \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


    $\displaystyle \int_0^{\infty} \frac{u^2}{u^4 + 1 } ~du = \frac{\pi}{2\sqrt{2}} $ this result helps evaluate the integral $\displaystyle \int_0^{\infty} \cos(x^2)~dx $ which can be started in a quite similar way .
    Last edited by simplependulum; Sep 12th 2010 at 02:08 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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