Solution
= Solution
{c}
The absolute value is bounded by $e^{-ax}$, whose integral is $1/a$. Thus the <function> is <Lebesgue integrable>. The complex integral of an <exponential function> is
$$
\int_0^\infty e^{(-a+ib)x}\,dx=\left[\frac{e^{(-a+ib)x}}{-a+ib}\right]_0^\infty=\frac1{a-ib}=\frac{a+ib}{a^2+b^2}.
$$
Taking the <real part>, justified by <absolute integrability>, gives
$$
\boxed{\int_0^\infty e^{-ax}\cos(bx)\,dx=\frac a{a^2+b^2}.}
$$