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Math Help - Definite Integral, infinite limits, complex analysis

  1. #1
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    Definite Integral, infinite limits, complex analysis

    I need to show that:

    \displaystyle{\int_{0}^{\infty}\frac{\sin{x}}{x}\,  dt=\frac{1}{2i}\int_{-\infty}^{\infty}\frac{e^{ix}-1}{x}}

    It seems obvious that i need to use

    \sin{x}=\frac{e^{ix}-e^{-ix}}{2i}

    but I can't get it to work.

    I have tried thinking it of as a complex curve integral

    \displaystyle{\int_{\gamma}^{}f(z)\,dz \ \,\ \ f(z)=\frac{\sin{z}}{z}

    with

    \gamma(t)=t, 0<t<\infty

    and trying to use a different parametrization for \gamma(t)
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  2. #2
    A Plied Mathematician
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    I'd be willing to bet that you could take the RHS, substitute exp(ix) = cos(x) + i sin(x), and then use properties of even and odd functions to get the desired result. Are you sure that that -1 is in the numerator of the RHS's integrand? Because that's not such a nice feature of the problem.
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  3. #3
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    Yes, the -1 is definitely there. It is straight out of the book.

    Maybe it has something to do with the limit of \frac{\cos{x}-1}{x} as x -> 0

    Edit: Your idea does seem to work since \frac{\cos{x}-1}{x} is odd and should integrate to zero. I was trying to go from the left and show it equals the right though which was obviously the wrong way. Cheers!
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  4. #4
    A Plied Mathematician
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    Glad you got it to work. Have a good one!
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