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.
 A: Looking at derivatives of the functions $\sin(\alpha t) e^{-\lambda t}$ and $\cos(\alpha t) e^{-\lambda t}$, it's reasonable to try for an antiderivative of the form
$$ F(t) = (A \sin(\alpha t) + B \cos(\alpha t)) e^{-\lambda t} $$
where $A$ and $B$ are constants.
Differentiating this, 
$$ F'(t) = ((-A \lambda - B \alpha) \sin(\alpha t) + (A \alpha - B \lambda) \cos(\alpha t)) e^{-\lambda t} $$
So this will be correct if
$$ \eqalign{-A\lambda - B \alpha &= 0\cr
             A \alpha - B \lambda &= 1\cr}$$
Solve this system of equations for $A$ and $B$...
A: 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. 
A: 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?
A: Note that
$$\frac d{dt}\frac{-\lambda\cos(\alpha t)e^{-\lambda t}+\alpha\sin(\alpha t)e^{-\lambda t}}{\alpha^2+\lambda^2}=\cos(\alpha t)e^{-\lambda t}$$
And apply the fundamental theorem of calculus.
A: Integrate by parts twice, each time either integrating the trig function or the exponential.  Then "solve for the integral" and the rest is easy.  See if you can answer your own question this way.
