I'm trying to do an exercise in Folland (8.45), and I'm having to calculate $W_t=\left[\frac{\sin (2\pi t|\xi|)}{2\pi|\xi|}\right]^\vee,$ where $\vee$ is the inverse Fourier transform. We can make free use of the identity: $ \chi_{[-a,a]}^\vee(x) = \frac{\sin(2\pi a x)}{\pi x },$
I must have some elementary confusions.
Here's my attempt:
$\begin{eqnarray*} W_t&=&\int_\mathbb{R} \frac{\sin (2\pi t|\xi|)}{2\pi|\xi|}\cdot e^{-2\pi i |\xi|x}\ dx \\ &=&\frac12 \int_\mathbb{R} \chi_{[-t,t]}^\vee(|\xi|)\cdot e^{-2\pi i |\xi|x}\ dx \\ &=&\frac12 \int_\mathbb{R} \left(\int_\mathbb{R} \chi_{[-t,t]}(x)\cdot e^{-2\pi i |\xi| y}\ dy\right)\cdot e^{-2\pi i |\xi|x}\ dx \\ &=& \frac12 \int_\mathbb{R} \left(\int_{-t}^t e^{-2\pi i |\xi| y}\ dy\right)\cdot e^{-2\pi i |\xi|x}\ dx \\ &=& \frac12 \int_\mathbb{R} \left(\left(\frac{1}{2\pi i|\xi|}\right)\left(e^{2\pi i t}-e^{-2\pi i t}\right)\right)\cdot e^{-2\pi i |\xi|x}\ dx \\ &=& \frac{(e^{2\pi i t}-e^{-2\pi i t})}{4\pi i}\int_\mathbb{R} \frac{e^{-2\pi i |\xi|x}}{|\xi|}\ dx, \end{eqnarray*} $
which doesn't seem to converge...what am I doing wrong?
If it matters, the bigger question is to show (using the above identity) that $u(x,t):=f*\partial_t W_t(x)+g*W_t(x)=\frac12[f(x+t)+f(x-t)]+\frac12\int_{x-t}^{x+t}g(s)\ ds,$ subject to the initial conditions $(\partial_t^2-\Delta)u=0,\ \ u(x,0)=f(x), \ \ \partial_t u(x,0)=g(x).$