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Math Help - Sine squared integral

  1. #1
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    Sine squared integral

    Evaluate \int_0^\infty\left(\frac{\sin x}x\right)^2\,dx
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  2. #2
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    Use complex analysis, the answer is \frac{\pi}{2}.

    By considering the function f(z) = \frac{\frac{1}{2i}(z-z^{-1})}{z}.

    This is Mine 77th Post!!!
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  3. #3
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    Quote Originally Posted by liyi View Post
    Evaluate \int_0^\infty\left(\frac{\sin x}x\right)^2\,dx
     <br />
\int_0^\infty {\left( {\frac{{\sin x}}{x}} \right)^2 \,dx} = \int_0^\infty {\frac{{1 - \cos 2x}}{{2x^2 }}\,dx} .

    Use the following parameter: \frac{1}{{x^2 }} = \int_0^\infty {ue^{ - ux} \,du} .

    (The only reason that I'd ever create a double integral is that I can reverse the integration order.)

    From there you can solve the rest, without problems.
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  4. #4
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    Or we can define

    I(\varphi ) = \int_0^\infty {\frac{{\sin ^2 (\varphi x)}}<br />
{{x^2 }}\,dx} ,\,\varphi > 0.

    Then you can apply the Leibniz's Rule for differentiation under the integral sign. You'd get

    I(\varphi ) = \frac{\pi}<br />
{2}\varphi .

    The conclusion follows.
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  5. #5
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    \int_{0}^{\infty} \frac{\cos ax - \cos bx}{x^2} dx = \int_{0}^{\infty} \int_a^b \frac{\sin tx}{x}dt~dx = \int_a^b \int_0^{\infty}\frac{\sin tx}{x}dx~dt.

    But \int_0^{\infty} \frac{\sin tx}{x} dx = \int_a^b \frac{\pi}{2} dt = \frac{\pi}{2}(b-a).

    That means,
    \int_0^{\infty} \frac{\sin^2 x}{x^2} dx = \frac{1}{2}\int_0^{\infty}\frac{1-\cos 2x}{x^2} dx = \frac{1}{2} \int_0^{\infty} \frac{\cos 0x - \cos 2x}{x^2} dx = \frac{1}{2}(2\pi ) = \pi.

    Where did I make a mistake? Anyways, you get the idea I am too lazy to check this right now.
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