Lebesgue integration question

**charlotte12** · March 8th 2010, 03:11 PM

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!

**Opalg** · March 9th 2010, 03:27 AM

Originally Posted by charlotte12

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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