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Math Help - Complex Indented Paths to Solve Real Trig Integrals

  1. #1
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    Complex Indented Paths to Solve Real Trig Integrals

    Hello!

    How can someone solve this? I believe the answer is pi/2 but I don't know how to show this. Could someone walk me through it?

    \int_{0}^{\infty} \frac{sin^2x}{x^2} dx

    Thanks for your help! This question has me really stumped.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JoAdams5000 View Post
    Hello!

    How can someone solve this? I believe the answer is pi/2 but I don't know how to show this. Could someone walk me through it?

    \int_{0}^{\infty} \frac{sin^2x}{x^2} dx
    Thanks for your help! This question has me really stumped.
    Do you have to solve this using complex analysis?

    If so,
    Spoiler:

    Let's consider \displaystyle f(\zeta)=\frac{1-e^{2i\zeta}}{\zeta^2}. Then, clearly f is analytic except it has a simple pole at 0. So, we may define a family of contours C\left(\varepsilon,R\right),\quad 0<\varepsilon<R given by C\left(\varepsilon,R\right)=\Gamma_R\cup\gamma_{\v  arepsilon}\cup [-R,\varepsilon]\cup[\varepsilon,R] where \Gamma_R is the semi-circle on the x-axis going through -R,iR,R and \gamma_\varepsilon is the semi-circle on the x-axis passing through -\varepsilon,i\varepsilon,\varpepsilon and [-R,-\varepsilon],[\varepsilon,R] are just the intervals on the real line; of course orient C\left(\varepsilon,R\right) counterclockwise. Note then by the Residue Theorem that

    \displaystyle 0=\int_{C\left(\varepsilon,R\right)}f(\zeta)d\zeta  =\int_{\Gamma_R}f(\zeta)d\zeta+\int_{\gamma_{\vare  psilon}}f(\zeta)d\zeta+\int_{[-R,\varepsilon]}f(\zeta)d\zeta+\int_{[R,\varepsilon]}f(\zeta)d\zeta

    Evidently it's true that

    \displaystyle |f(z)|=\frac{\left|1-e^{2iz}\right|}{z^2}\leqslant \frac{2}{|z|^2}

    and so it easily follows that

    \displaystyle \int_{-\infty}^{\infty}\frac{1-e^{2i\zeta}}{\zeta^2}d\zeta=\pi \text{Res}\left(f(\zeta),0\right)=\pi i(-2i)=2pi

    but the whole point to picking f(\zeta) was that \sin^2(\zeta)=\frac{1}{2}\text{Re}\left(1-e^{2i\zeta}\right) and thus

    \displaystyle \int_{-\infty}^{\infty}\frac{\sin^2(x)}{x^2}dx=\frac{1}{4  }\text{Re}\int_{-\infty}^{\infty}\frac{1-e^{2i\zeta}}{\zeta^2}d\zeta=\frac{\pi}{2}

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