Laplace transform of $\cos(at)/t$ 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
 A: 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.
A: 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. 
$$\begin{align}
F(t) &&= L^{-1} \log(s^2 + a^2)\\ 
\implies tF(t) &&= L^{-1}-\frac{d}{ds}log(s^2 + a^2)\\
\implies tF(t) &&= L^{-1}-\frac{2s}{s^2 +a^2}\\
\implies tF(t) &&= -2\cos(at) \\ 
\implies F(t) &&= -2\frac{\cos(at)}{t}\\
\implies \frac{\cos(at)}{t} &&= \frac{F(t)}{-2}\\
\implies L\frac{\cos(at)}{t} &&= L(\frac{F(t)}{-2})\\ 
\implies L\frac{\cos(at)}{t} &&= \frac{\log|s^2 + a^2|}{-2})\\
\end{align}$$
