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Math Help - Please help me derive this value of the integral using an already derived result!

  1. #1
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    Wah Cantt
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    Post Please help me derive this value of the integral using an already derived result!

    By the SINE INTEGRAL OF SINE PULSE, we derived in the class that
    pi/2 = integral 0 to infinity ( ( sin w ) / w ) dw

    now PLEASE PROVE that "" integral 0 to infinity ( (sin^2 w ) / w^2 ) dw = pi/2 "" as well.

    I'll be really very thankful as it is very urgent. Thanks
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  2. #2
    Newbie russo's Avatar
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    Re: Please help me derive this value of the integral using an already derived result!

    Hi. I'm not sure how you're supposed to solve it but, using Fourier transform (I assumed that this could be the way since you had a function against \omega) I arrived to the result:

    First, let's call your function sinc(\omega) = \frac{sin(\omega)}{\omega} so the integral you want to solve is \int_{0}^{\infty}sinc^2(\omega) \,d\omega

    Then, from Parseval's identity:

    (1) \int_{-\infty}^{\infty}|f(t)|^2\,dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}|F(\omega)|^2\,d\omega

    We can assume that (2) F(\omega) = \left\{\begin{array}{c l}  0 & \mbox{ if } \omega < 0 \\  sinc^2(\omega) & \mbox{ if } \omega \geq 0\end{array}\right.

    Since we have F(\omega), we might want to find f(t) to solve it easily. First, I'll rewrite the function so we can use a transform I've got from my table.

    sinc(\omega) = \frac{2}{2}sinc(\omega)

    The trasform is: P_a(t) \Leftrightarrow a\cdot sinc\left(\frac{a\omega}{2}\right). Where P stands for square pulse and a total width.

    Antitransforming:

    (3) \mathcal{F}^{-1}\left\{\frac{2 \cdot sinc(\omega)}{2}\right\} = \frac{P_2(t)}{2}

    Once we have everything, we can assume from (1), (2) and (3) that:

    2\pi \cdot \int_{0}^{\infty}\frac{P_2\,^2(t)}{2^2} \,dt = \int_{0}^{\infty}sinc^2(\omega)}\,d\omega

    Now we solve the first part (which is easier):

    2\pi \cdot \int_{0}^{\infty}\frac{P_2\,^2(t)}{4} \,dt = \frac{\pi}{2}\int_{0}^{1}\,dt = \frac{\pi}{2} \cdot \left. t \right |_0^1 = \frac{\pi}{2}

    I hope this is what you needed.
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