improper integrals - Evaluating $int_0^{infty} frac{sin(xt)(1-cos(at))}{t^2} dt$

The problem is to evaluate the improper integral:
$I = \int_0^{\infty} \frac{\sin(xt)(1-\cos(at))}{t^2} dt$.

This can be written as follows:

$$I = \int_0^{\infty} dt \frac{\sin(xt)}t \int_0^a \sin(yt)dy = \int_0^{\infty} dt \int_0^a \frac{\sin(xt)\sin(yt)}t dy$$

In a previous problem, I determined:

$$J = \int_0^{\infty} \frac{\sin(xt)\sin(yt)}t dt = \frac12\log\left(\lvert\frac{x-y}{x+y}\rvert\right)$$
which converges uniformly with respect to $y$ as long as $|x| \ne |y|$. Hence the order of integration can be interchanged for $I$ to get:

$$I = \int_0^a dy \int_0^{\infty} \frac{\sin(xt)\sin(yt)}t dt = \int_0^a \frac12\log\left(\lvert\frac{x-y}{x+y}\rvert\right)dy$$

This result may not be valid if $|a| \ge |x|$ since $J$ becomes unbounded if $|x|=|y|$. If I integrate $J$ with respect to $y$ from $0$ to $a$, $y$ is going to eventually reach $|x|$ or $-|x|$ if $|a| \ge |x|$.

Is it valid to integrate with respect to $y$ over singularities in $J$ provided that $\int_0^a \frac12\log\left(\lvert\frac{x-y}{x+y}\rvert\right)dy$ converges? Or does $I$ actually diverge if $|a| \ge |x|$?

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I think the proper way to do this is as follows. Assume $a > x > 0$ to simplify the problem. Then I should integrate as follows:

$$I = \int_0^{\infty} dt \int_0^{x-\epsilon} \frac{\sin(xt)\sin(yt)}t dy + \int_{x-\epsilon}^{x+\epsilon} \frac{\sin(xt)\sin(yt)}t dy + \int_{x+\epsilon}^a \frac{\sin(xt)\sin(yt)}t dy$$

Then I must show that the middle integral approaches $0$ and $\epsilon \rightarrow 0$. But the outside integrals diverge.