Solution to this integral? could anyone solve this integral ?
$$\int_0^\infty \frac{e^{-x}\sin(x)\cos(ax)}x~\mathrm dx$$
well i have tried opening up the sin*cos using trigonometric identities but that didn't help so much

By the sine addition formulas, it is enough to compute
$$ f(m)=\int_{0}^{+\infty}\frac{\sin(mx)}{x}e^{-x}\,dx \tag{1}$$
where $f(0)=0$ and by the dominated convergence theorem
$$ f'(m) = \int_{0}^{+\infty}\cos(mx)e^{-x}\,dx\stackrel{IBP}{=}\frac{1}{1+m^2}\tag{2}$$
implying:
$$ f(m)=\int_{0}^{+\infty}\frac{\sin(mx)}{x}e^{-x}\,dx = \arctan(m)\tag{3} $$
and
$$\begin{eqnarray*} \int_{0}^{+\infty}\frac{\sin(x)\cos(ax)}{x}e^{-x}\,dx &=& \frac{\arctan(1-a)+\arctan(1+a)}{2}\\&=&\color{red}{\frac{1}{2}\,\arctan\frac{2}{a^2}}.\tag{4} \end{eqnarray*}$$
