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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!

Elmo
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  • Notice that $1/s$ is the Laplace transform of the Heaviside step function. If you wish to shift the jump in this function to $s$ (where your integral starts), you would have an integrand of $\frac{1}{s}\mathrm{e}^{- s^2}$, not just $1/s$. – Eric Towers Oct 07 '16 at 06:52
  • the definition of the Laplace transform is $L[f(t)] = \int_0^\infty f(t)e^{-st}dt$. And $L[\frac{1}{t}] = \int_0^\infty \frac{1}{t}e^{-st}dt$ diverges for every $s$ – reuns Oct 07 '16 at 06:53
  • @EricTowers your 1st comment is impossible to understand. And the (bilateral) Laplace transform of $\frac{1_{t >\ -1}}{t}$ exists in the sense of distribution or principal value – reuns Oct 07 '16 at 06:55
  • @user1952009 : The function whose Laplace transform is $1/s$ is not $1$, it is the unit step function (with step at $0$). This is unambiguously not a bilateral Laplace transform. It is also not the unilateral Laplace transform unless $s=0$, which causes problems in OP's second display. It is therefore, one of the relatively common variants that allow imprecisely specified integration on the interval $[s,\infty)$ for some $s>0$. However, this means that OP's "$1$" is not the constant function $1$, but is the unit step function with step at $s$, which Laplace transform is the one I give. – Eric Towers Oct 07 '16 at 07:01
  • Re: your edit: Can you show that the part of the integral over $[0,1]$ diverges? – Eric Towers Oct 07 '16 at 07:04
  • @EricTowers what you write is impossible to understand. $1/t$ is completely different to $1/s$, since $s$ is complex and it would be $f(t) = \frac{1}{\sigma+i t}$ whose (unilateral) Laplace transform is well-defined for $\sigma \ne 0$ – reuns Oct 07 '16 at 07:05
  • Let us continue this discussion in chat. – Eric Towers Oct 07 '16 at 07:08
  • @Elmo write that $e^{-st} = 1+ (e^{-st}-1)$ and $|e^{-st}-1| < 2|s|t$ as $t \to 0$, therefore $\int_0^1 (e^{-st}-1) \frac{1}{t}dt$ converges, and $\int_0^1 e^{-st} \frac{1}{t}dt = \color{red}{\int_0^1 \frac{1}{t}dt }+\int_0^1 (e^{-st}-1) \frac{1}{t}dt$ – reuns Oct 07 '16 at 07:09
  • @user1952009 : If you have an Answer, there is a perfectly good box, labelled "Your Answer", to contain it. Comments are not for answers. – Eric Towers Oct 07 '16 at 07:13
  • @EricTowers you are funny. The problem is that what Elmo wrote is impossible to understand (things like $\int_s^\infty F(t)dt$) – reuns Oct 07 '16 at 07:16
  • @user1952009 : Incorrect. What Elmo wrote is entirely common in engineering and other sciences. That you are not familiar with those uses does not automatically invalidate them. – Eric Towers Oct 07 '16 at 07:16
  • @user1952009 : In fact, the transform pair you are calling out has a name in the literature. It is "frequency domain integration". It's line 7 in the table of pairs here. – Eric Towers Oct 07 '16 at 07:18
  • I don't see what you mean, at all. And tks I know all my Fourier transforms – reuns Oct 07 '16 at 07:21
  • @user1952009 : Evidently not. – Eric Towers Oct 07 '16 at 07:22
  • @EricTowers ?? ${}{}$ – reuns Oct 07 '16 at 07:26

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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.

Eric Towers
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