Solving the improper integral $\int_0^{\infty} {e^{-ax}\cos{(bx)}} dx$ 
$$\int_0^{\infty} {e^{-ax}\cos{(bx)}} dx$$

I know I need to use integration by part method. But I'm not sure how does the improper integral take place?

Just an idea:
$$\cos bx=\frac{e^{ibx}+e^{-ibx}}{2}\implies \int\limits_0^\infty e^{-ax}\cos bx\,dx=\frac{1}{2}\int\limits_0^\infty\left(e^{-(a-ib)x}+e^{-(a+ib)x}\right)dx=\ldots$$
Check the values of $\,a,b\,$ (reals, I presume...?) for which the above converges.
