A question regarding the computation of $\mathcal{L}_s[f'(t)/t]$, where $f(t)$ is a differentiable function, was asked few hours ago. Unfortunately, this question was voluntarily deleted by the OP.
I finally decided to answer this question just for mere curiosity. Firstly, I noted that, from the definition of Laplace transform of a function $f'(t)$:
$$ G(s) = \mathcal{L}_sf'(t) = \int^\infty_0 f'(t) e^{-st} \, \mathrm{d}t,$$
we can integrate this function of $s$ with respect to $s$ to have:
$$\int G(s)\mathrm{d}s = -\int^\infty_0 \frac{f'(t)}{t}e^{-st} \, \mathrm{d}t - K = - \mathcal{L}[f'(t)/t]-K,$$
where $K$ is a constant of integration. But, since $G(s) = s F(s)-f(0)$, where $F(s)$ is the Laplace transform of $f(t)$, we ultimately come up with:
$$\color{blue}{ \mathcal{L}_s[f'(t)/t]= C - \int sF(s) \mathrm{d}s+sf(0) }$$
where $C$ is a new constant of integration.
Then, I got two questions out of this:
- Is the value of $C$ undetermined? Can I set it to $0$? Otherwise, how can I determine it?
- Is it safe to integrate with respect to $t$ when the limits of integration contain $t=0$?
Any hints will be appreciated.
Cheers!
First call:
$$ G(u) = \mathcal{L}_uf'(t) = \int^\infty_0 f'(t) e^{-ut} \, \mathrm{d}t,$$
Integrate respect to $u$ from $0$ to $s$:
$$\int_0^s G(u)\mathrm{d}u = \int^\infty_0 \int_0^s f'(t)e^{-ut} \mathrm{d}u \, \mathrm{d}t$$ $$=\int^\infty_0 \frac{f'(t)}{t}[1-e^{-st}] \, \mathrm{d}t=- \mathcal{L}\left[\frac{f'(t)}{t}\right]+\int^\infty_0 \frac{f'(t)}{t} \, \mathrm{d}t$$
Replacing $G(u) = u F(u)-f(0)$:
$$\int_0^s (uF(u)-f(0))du=- \mathcal{L}\left[\frac{f'(t)}{t}\right]+\int^\infty_0 \frac{f'(t)}{t} \, \mathrm{d}t$$
Leads to:
$$\mathcal{L}\left[\frac{f'(t)}{t}\right]=\int_0^s (f(0)-uF(u))du+\int^\infty_0 \frac{f'(t)}{t} \, \mathrm{d}t$$
$$\mathcal{L}\left[\frac{f'(t)}{t}\right]=sf(0)-\int_0^s uF(u)du+\int^\infty_0 \frac{f'(t)}{t} \, \mathrm{d}t$$