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.

Explicitly, with the choice $$u = \cos \alpha t, \quad du = -\alpha \sin \alpha t \, dt, \\ dv = e^{-\lambda t}, \quad v = -\lambda^{-1} e^{-\lambda t},$$ the first integration by parts gives
$$I = \int e^{-\lambda t} \cos \alpha t \, dt = -\frac{1}{\lambda} e^{-\lambda t} \cos \alpha t - \frac{\alpha}{\lambda} \int e^{-\lambda t} \sin \alpha t.$$  Then, the second integration by parts with the choice $$u = \sin \alpha t, \quad du = \alpha \cos \alpha t \, dt, \\ dv = e^{-\lambda t}, \quad v = -\lambda^{-1} e^{-\lambda t},$$ gives $$I = -\frac{1}{\lambda}e^{-\lambda t} \cos \alpha t + \frac{\alpha}{\lambda^2} e^{-\lambda t} \sin \alpha t - \frac{\alpha^2}{\lambda^2} \int e^{-\lambda t} \cos \alpha t \, dt.$$  But this integral on the RHS is simply $I$, so we conclude $$\left(1 + \frac{\alpha^2}{\lambda^2} \right) I = \frac{e^{-\lambda t}}{\lambda^2} \left(\alpha \sin \alpha t - \lambda \cos \alpha t\right),$$ which gives the value of the indefinite integral as $$I = \frac{e^{-\lambda t}}{\lambda^2 + \alpha^2} (\alpha \sin \alpha t - \lambda \cos \alpha t) + C,$$ for some constant of integration $C$.  Taking the limit as $t \to \infty$, noting that $\sin \alpha t$ and $\cos \alpha t$ are bounded and finite, gives $$\int_{t = 0}^\infty e^{-\lambda t} \cos \alpha t \, dt = I(\infty) - I(0) = 0 - \left(- \frac{\lambda}{\lambda^2 + \alpha^2} \right)  = \frac{\lambda}{\lambda^2 + \alpha^2},$$ as claimed. 
