Show that the inverse Laplace transform of F(s) = logs is given by f(x) = -1/x

Question

Here's what I did, but I'm not sure if it's right/ valid.
F(s) = log(s)
L(xf(x)) = -1
So, L(xf(x)) = (-1)(1/s)
Therefore, xf(x) = -1
So, f(x) = -1/x

Answer 1

$\mathcal L^{-1}[F'(s)] = -x \mathcal L^{-1} [F(s)] = -x f(x)$

or $$\mathcal L^{-1} [F(s)] = \frac{-1}{x} \mathcal L^{-1}[F'(s)]$$ Let,$F(s) = log s$. So, $F'(s) = \frac{1}{s}$.

Thus

$$\mathcal L^{-1} [F(s)] = \mathcal L^{-1} [log(s)] = \frac{-1}{x} \mathcal L^{-1}[\frac{1}{s}] = \frac{-1}{x}.1= \frac{-1}{x} = f(x)$$

Answer 2

Let $f_n(x) = \frac{-1}{x} 1_{x > 1/n} + C_n \delta(x)$ where $C_n = \int_{1/n}^\infty \frac{e^{-x}}{x}dx$ and $\delta$ is the Dirac delta distribution satisfying $\mathcal{L}[\delta(x)](s) = 1$. Then $$\lim_{n \to \infty}\mathcal{L}[f_n(x)](s) =\lim_{n \to \infty}\int_{1/n}^\infty \frac{-1}{x}e^{-sx} dx+C_n= \lim_{n \to \infty} \int_{1/n}^\infty \frac{e^{-x}-e^{-sx}}{x} dx\\ = \int_0^\infty \frac{e^{-x}-e^{-sx}}{x} dx= \int_0^\infty \int_1^s e^{-zx}dzdx = \int_1^s\int_0^\infty e^{-zx}dxdz=\int_1^s \frac{1}{z}dz= \log s$$

Thus the inverse Laplace transform of $\log s$ is $\lim_{n \to \infty} f_n(x)$ with the limit taken in the sense of distributions. Concretely it means if $G(s) = \mathcal{L}[g(x)](s)$ with $g$ integrable and $C^1$ then $G(s) \log s = \mathcal{L}[h(x)](s)$ where $h(x) = \lim_{n \to \infty} g \ast f_n(x)$ (convolution).

Also $\lim_{n\to\infty} f_n$ is the distributional derivative of $(\gamma-\log x)1_{x > 0}$

Answer 3

It is more convenient to define $t_+^{-1}$ as the distributional derivative of $(\ln t)_+$. Then, since $\mathcal L[f'](s) = s \mathcal L[f](s)$ holds for distributions, $$\mathcal L[t_+^{-1}] = s \mathcal L[(\ln t)_+] = -\gamma - \ln s, \\ \mathcal L^{-1}[\ln s] = -t_+^{-1} - \gamma \delta(t).$$