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Math Help - an integral

  1. #1
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    Smile an integral

    Hi!

    I need the following to complete a proof.

    \int _0^\infty \frac{sin^2(x)}{x^2}dx = \frac{\pi}{2}

    I've checked it in Maple and it's true. All I have left is to actually show it (without referring to Maple ofc).

    Thanks!
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  2. #2
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    I expect you would have to use the residue theorem...
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by mgarson View Post
    Hi!

    I need the following to complete a proof.

    \int _0^\infty \frac{sin^2(x)}{x^2}dx = \frac{\pi}{2}

    I've checked it in Maple and it's true. All I have left is to actually show it (without referring to Maple ofc).

    Thanks!
    A solution of the integral based on the Laplace Tranform is illustrated here...

    http://www.mathhelpforum.com/math-he...rm-170211.html

    Kind regards

    \chi \sigma
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  4. #4
    Behold, the power of SARDINES!
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    Here is a third method using the Laplace transform

    Consider the function

    f(t)=\int_{0}^{\infty}\frac{\sin^2(tx)}{x^2}dx

    Note that

    f(1)=\int_{0}^{\infty}\frac{\sin^2(x)}{x^2}dx

    Is the integral you are looking for

    Using a trig Identity we get

    f(t)=\int_{0}^{\infty}\frac{\sin^2(tx)}{x^2}dx= \int_{0}^{\infty} \frac{1-\cos(2tx)}{2x^2}

    Now if we take the Laplace transform with respect to t we get

    \mathcal{L}(f)=\frac{1}{2}\int_{0}^{\infty} \left( \frac{1}{x^2s}-\frac{s}{x^2(s^2+4x^2)} \right) dx

    Now by partial fractions on the 2nd term we get

    \frac{s}{x^2(s^2+4x^2)}=\frac{1}{sx^2}-\frac{4}{s(s^2+4x^2)}

    Plugging this back in gives


    \mathcal{L}(f)=\frac{2}{s}\int_{0}^{\infty} \frac{1}{(s^2+4x^2)} dx=\frac{1}{s^2} \cdot \tan^{-1}\left( \frac{2x}{s}\right) \bigg|_{0}^{\infty}

    \mathcal{L}(f)=\frac{\pi }{2s^2}

    Now finally take the inverse transform to get

    f(t)=\frac{\pi}{2}t \implies f(1)=\frac{\pi}{2}
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Another way: using integration by parts with u=\sin^2 x,\; dv=dx/x^2 :

    \displaystyle\int_0^{+\infty}\dfrac{\sin ^2 x}{x^2}dx=\left[\dfrac{\sin ^2 x}{x}\right]_0^{+\infty}+\displaystyle\int_0^{+\infty}\dfrac{\  sin 2x}{x}dx=\displaystyle\int_0^{+\infty}\dfrac{\sin 2 x}{x}dx


    Using the substitution t=2x we obtain the Dirichlet's integral:

    \displaystyle\int_0^{+\infty}\dfrac{\sin 2 x}{x}dx=\displaystyle\int_0^{+\infty}\dfrac{\sin  t}{t}dt=\dfrac{\pi}{2}
    Last edited by FernandoRevilla; May 16th 2011 at 03:38 PM.
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