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Math Help - Lebesgue integration question

  1. #1
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    Lebesgue integration question

    Hello,

    For one of the integration questions I have to do, I need to show that

    f_{t}(x)=\frac{e^{-x}(1-\cos(tx))}{x^{2}}\mathbf{1}_{[0,\infty)}(x)

    is Lebesgue integrable, where 1 is the indicator function, 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}).

    Does anyone have any ideas about how I would solve this? Thanks for any help!
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  2. #2
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    Quote Originally Posted by charlotte12 View Post
    Hello,

    For one of the integration questions I have to do, I need to show that

    f_{t}(x)=\frac{e^{-x}(1-\cos(tx))}{x^{2}}\mathbf{1}_{[0,\infty)}(x)

    is Lebesgue integrable, where 1 is the indicator function, 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}).

    Does anyone have any ideas about how I would solve this? Thanks for any help!
    To see that f_t(x) is Lebesgue-integrable, show that |f_t(x)| is dominated by some multiple of e^{-x}. (Notice that f_t(x) is bounded as x\searrow0.)

    To calculate the integral, let g(t) = \int_{0}^{\infty}\frac{e^{-x}(1-\cos(tx))}{x^{2}}dx. Differentiate twice under the integral sign to see that g''(t) = \int_0^\infty\!\!\! e^{-x}\cos(tx)dx. You can evaluate that integral (integration by parts twice) to get g''(t) = \frac1{1+t^2}. Then integrate twice, using the fact that g'(0) = g(0) = 0 to get the constants of integration, and you will find that g(t) = t\tan^{-1}t-\tfrac{1}{2}\log(1+t^{2}).
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