Lebesgue integration question

• Mar 8th 2010, 03:11 PM
charlotte12
Lebesgue integration question
Hello,

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

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

is Lebesgue integrable, where 1 is the indicator function, and that

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

Does anyone have any ideas about how I would solve this? Thanks for any help!
• Mar 9th 2010, 03:27 AM
Opalg
Quote:

Originally Posted by charlotte12
Hello,

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

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

is Lebesgue integrable, where 1 is the indicator function, and that

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

Does anyone have any ideas about how I would solve this? Thanks for any help!

To see that $\displaystyle f_t(x)$ is Lebesgue-integrable, show that $\displaystyle |f_t(x)|$ is dominated by some multiple of $\displaystyle e^{-x}$. (Notice that $\displaystyle f_t(x)$ is bounded as $\displaystyle x\searrow0$.)

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