I know that $f(t)=\frac{1}{t}$ does not have a Laplace transformation but I don't know the right(formal) way to prove it.
Here is my way:
Using the fact:
$L[\frac{f(t)}{t}]=\int_{s}^{\infty}F(t)dt$,
I have:
$L[\frac{1}{t}]=\int_{s}^{\infty}\frac{1}{s}dt = \frac{1}{s}|^{\infty}_{s}$
this is divergent and therefore the Laplace transformation of $f(t)=\frac{1}{t}$ does not exist.
However, I think my way is not rigorous. I hope I can receive a explicit proof of it (Maybe by definition to integrate).
Thank you so much!
Edit:
Actually, I do not know how to actually prove that $\int_{0}^{\infty}e^{-st}\frac{1}{t}dt$ diverges. Please help me with this!
We show that $\int_0^\infty \mathrm{e}^{-st} \frac{1}{t} \,\mathrm{d}t$ diverges. Since this integrand is everywhere positive, it is sufficient to show that $\int_0^1 \mathrm{e}^{-st} \frac{1}{t} \,\mathrm{d}t$ diverges. Note that for $t \in [0,1]$ and $s \geq 0$, $\mathrm{e}^{st} \leq \mathrm{e}^{s}$ and so $\mathrm{e}^{-st} \geq \mathrm{e}^{-s}$. Consequently, $$\int_0^1 \mathrm{e}^{-st} \frac{1}{t} \,\mathrm{d}t \geq \int_0^1 \mathrm{e}^{-s} \frac{1}{t} \,\mathrm{d}t = \mathrm{e}^{-s}\int_0^1 \frac{1}{t} \,\mathrm{d}t \text{,} $$ and we recognize this latter integral as one which diverges to infinity. Therefore, the Laplace transform of $\frac{1}{t}$ (with respect to $t$) does not exist.