1. ## Help with integration

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.

2. Originally Posted by Mimi89

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 $\int_0^t\tan^{-1}u\,du = t \tan^{-1} t - \tfrac{1}{2}\log (1+t^{2})$.