Using Laplace Transforms to Evaluate Integrals

Question

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

Answer 1

1. 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}

2. 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}