Using Laplace Transforms to Evaluate Integrals 
I'm trying to solve $$\int_0^{\infty} \, \frac{e^{-2t}\cos(3t)-e^{-4t}\cos(2t)}{t}dt.$$ 

I'm sure that this involves Laplace transforms, I'm just not sure how.
I would start by separating and making two integrals, where the second would solve in the same procedure as the first.  Now I would have 
$\int_{0}^{\infty} e^{-2t}cos(3t)t^{-1}dt$, where from here I'm not sure what to do.  Normally I would use something like $L{tf(t)}=-F'(s)$, but that doesn't apply here.  Any help is appreciated, thanks
 A: *

*Method 1


By using the integral
\begin{align}
\int_{0}^{\infty} e^{-u t} \, du = \frac{1}{t}
\end{align}
the integral
\begin{align}
I = \int_0^{\infty} \, \frac{e^{-2t}\cos(3t)-e^{-4t}\cos(2t)}{t}dt
\end{align}
becomes
\begin{align}
I &= \int_0^{\infty} \int_{0}^{\infty} (e^{-2t}\cos(3t)-e^{-4t}\cos(2t) ) ds dt \\
&= \int_{0}^{\infty} \, ds \, \left[ \int_{0}^{\infty} e^{-(s+2)t} \cos(3t) \, dt - \int_{0}^{\infty} e^{-(s+4)t} \cos(2t) \, dt \right] \\
&= \int_{0}^{\infty} \left[ \frac{s+2}{(s+2)^{2} + 3^{3}}  - \frac{s+4}{(s+4)^{2} + 4} \right] \, ds \\
&= \frac{1}{2} \left[ \ln\left( \frac{(s+2)^{2} + 9}{(s+4)^{2} + 4} \right) \right]_{0}^{\infty} \\
&= - \frac{1}{2} \, \ln\left( \frac{13}{20} \right).
\end{align}


*Method 2


Using the Laplace transform of a of a function divided by the variable is under the rule
\begin{align}
\mathcal{L}\left\{ \frac{f(t)}{t} \right\} = \int_{s}^{\infty} F(u) \, du
\end{align}
where $F(s)$ is the transformed function. Fron this rule it is seen that
\begin{align}
I &= \int_{2}^{\infty} \frac{u}{u^{2} + 3^{2}} \, du - \int_{4}^{\infty} \frac{u}{u^{2} + 2^{2}} \, du \\
&= \frac{1}{2} \left[ \ln(u^{2} + 9) \right]_{2}^{\infty} - \frac{1}{2} \left[ \ln(u^{2} + 4) \right]_{4}^{\infty} \\
&= \frac{1}{2} \, \lim_{u \rightarrow \infty} \left\{ \ln\left( \frac{u^{2}
+ 9}{u^{2} + 4} \right) \right\} - \frac{1}{2} \, \ln\left(\frac{2^{2} + 9}
{4^{2} + 4} \right) \\
&= \frac{1}{2} \, \lim_{u \rightarrow \infty} \left\{ \ln\left( \frac{1 + \frac{9}{u^{2}} }{ 1 + \frac{4}{u^{2}} } \right) \right\} - \frac{1}{2} \,
\ln\left( \frac{13}{20} \right) \\ 
&= - \frac{1}{2} \ln\left( \frac{13}{20} \right).
\end{align}
