Solution
= Solution
{c}
Again $|e^{-ax}\sin(bx)|\leq e^{-ax}$ proves <Lebesgue integrability>. Taking the <imaginary part> of the absolutely convergent complex integral in part <2/b/i> gives
$$
\boxed{\int_0^\infty e^{-ax}\sin(bx)\,dx=\frac b{a^2+b^2}.}
$$