How do i find the lapalace transorm of this intergral using the convolution theorem?

Question

$$\int_0^{t} e^{-x}\cos x \, dx$$

In the book, the $x$ is written as the greek letter "tau". Anyway, I'm confused about how to deal with this problem because the $f(t)$ is clearly $\cos t$, but $g(t)$ is not clear to me. Please help.

Answer 1

$f(t) = \int_0^t e^{-\tau} \cos \tau d \tau = e^{-t }\int_0^t e^{t-\tau} \cos \tau d \tau = (e^{-t })\left( (x \mapsto u(x)e^{x}) * (x \mapsto u(x)\cos x) \right)(t)$.

$({\cal L}f )(s) = ({\cal L}(x \mapsto u(x)e^{x}) )(s+1) ({\cal L}(x \mapsto u(x) \cos x) )(s+1) $. Note the $s+1$ parameter to deal with the multiplication by $e^{-t}$.

Answer 2

So the transform becomes $\frac{s}{s^2 + 1} \frac{1}{s-1}$ with the shift $s+1$.