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Math Help - Help with integration

  1. #1
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    Help with integration

    Could you please help me with this:

    show that for -T<t<T, the function  f_{t} : \mathbb{R} \rightarrow \mathbb{R} defined by
     f_{t} (x) = \frac{e^{-x} (1-cos(tx))}{x^2} \bold{1}_{ [0, \infty) } (x)

    is Lebesgue integrable and that

     \int_{0}^{\infty} \frac{e^{-x} (1-cos(tx))}{x^2} dx = t .tan^{-1} t - \frac{1}{2} log (1+t^{2})

    We are allowed to use results from Lebesgue integration.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Mimi89 View Post
    Could you please help me with this:

    show that for -T<t<T, the function  f_{t} : \mathbb{R} \rightarrow \mathbb{R} defined by
     f_{t} (x) = \frac{e^{-x} (1-cos(tx))}{x^2} \bold{1}_{ [0, \infty) } (x)

    is Lebesgue integrable and that

     \int_{0}^{\infty} \frac{e^{-x} (1-cos(tx))}{x^2} dx = t .tan^{-1} t - \frac{1}{2} log (1+t^{2})

    We are allowed to use results from Lebesgue integration.
    One way would be to use Fubini's theorem. Start with the fact that \int_0^\infty e^{-x}\cos(tx)\,dx = \frac1{1+t^2}. The justification for that is that e^{-x}\cos(tx) is dominated by e^{-x} and therefore integrable over [0,\infty); and the integral is equal to \text{Re}\!\int_0^\infty e^{-(1+it)x}dx = \text{Re}\left.\frac{e^{-(1+it)x}}{-(1+it)}\right|_0^\infty = \frac1{1+t^2}.

    Integrate with respect to t to see that \int_0^T\!\!\!\int_0^\infty e^{-x}\cos(tx)\,dxdt = \int_0^T\frac1{1+t^2}dt. Now use Fubini to change the order of integration and deduce that \int_0^\infty\frac{e^{-x}\sin(tx)}xdx = \tan^{-1}t (after replacing T by t).

    Now repeat the process: integrate from t=0 to T again, use Fubini, and conclude that  \int_{0}^{\infty} \frac{e^{-x} (1-cos(tx))}{x^2} dx = \int_0^t\tan^{-1}u\,du. Finally, integrate by parts to see that \int_0^t\tan^{-1}u\,du =  t \tan^{-1} t - \tfrac{1}{2}\log (1+t^{2}) .
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