How to prove that $\int_0^{\infty}\cos(\alpha t)e^{-\lambda t}\,\mathrm{d}t=\frac{\lambda}{\alpha^2+\lambda^2}$ using real methods? Could you possibly help me prove $$\int_0^{\infty}\cos(\alpha t)e^{-\lambda t}\,\mathrm{d}t=\frac{\lambda}{\alpha^2+\lambda^2}$$
I'm an early calculus student, so I would appreciate a thorough answer using real and as basic as possible methods. Thank you in advance.

Series method:
$$
E_n:=\int_0^\infty t^n e^{-\lambda t}dt=n!\lambda^{-n-1}
$$
and
$$
\sum_{n=0}^\infty \frac{(-1)^n a^{2n}t^{2n}}{(2n)!} = \cos(at)
$$
so
$$
\int_0^\infty \cos(at)e^{-\lambda t}dt = 
\sum_{n=0}^\infty \frac{(-1)^na^{2n}E_{2n}}{(2n)!}
=\sum_{n=0}^\infty (-1)^n a^{2n} \lambda^{-1-2n} =\frac{\lambda}{a^2+\lambda^2}
$$
But of course integration by parts is more elementary?
