If someone could help me solve for $$\mathcal{L}\left\{\frac{\cos(at)}{t}\right\}$$ it would be great.
Step-by-step I have so far:
$$\begin{align}\int_0^\infty \frac{\cos(at)\space e^{-st}}{t}\space\text{d}s &= \int_s^\infty \frac{s}{s^2+a^2}\space\text{d}s \\ &= \frac{1}{2} \int_s^\infty \frac{2s}{s^2+a^2}\space\text{d}s \\ \end{align}$$ Help me after this step please
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$\begingroup$Maybe I am misunderstanding something, but the integral should be in $t$, right? Ordinarily, if we have a function $f(t)$, the Laplace transform is $$(\mathcal Lf)(s) = \int^\infty_0 f(t) e^{-st} dt.$$ Here, because of the behavior near $t =0$, we see that the integral $$\int_0^\infty \frac{\cos(at)}{t} e^{-st} dt$$ doesn't converge. Indeed, for small $\varepsilon > 0$, we see by Taylor expansion that $$\int^\varepsilon_0 \frac{\cos(at)}{t} e^{-st} dt \approx \int^\varepsilon_0 \frac 1 t dt$$ which diverges.
$\endgroup$ 4 $\begingroup$Let us start with a fresh approach.
Here is a property of Laplace Transforms I'd like to use :
$$\begin{align} &&f(t) = L^{-1} F(s)\\ \implies&& tf(t) = L^{-1} -\frac{d}{ds} F(s) \\ \end{align}$$
So let's begin. Let $F(t)$ satisfy this equation. $$\begin{align} F(t) &&= L^{-1} \log(s^2 + a^2)\\ \implies tF(t) &&= L^{-1}-\frac{d}{ds}log(s^2 + a^2)\\ \implies tF(t) &&= L^{-1}-\frac{2s}{s^2 +a^2}\\ \implies tF(t) &&= -2\cos(at) \\ \implies F(t) &&= -2\frac{\cos(at)}{t}\\ \implies \frac{\cos(at)}{t} &&= \frac{F(t)}{-2}\\ \implies L\frac{\cos(at)}{t} &&= L(\frac{F(t)}{-2})\\ \implies L\frac{\cos(at)}{t} &&= \frac{\log|s^2 + a^2|}{-2})\\ \end{align}$$
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