Does the laplace transform of $1/t$ exist? If yes, how do we calculate it? Putting it in $$\int_0^\infty (e^{-st}/t) dt$$ won't solve.

Is there any other way? If not, why?


  • 1
    $\begingroup$ Does this exist? No, for the reason you quote. $\endgroup$ – Did May 19 '12 at 19:12
  • $\begingroup$ If that is the definition of "Laplace transform", then (as you note) it diverges. What do you want it for? Maybe something else will do... $\endgroup$ – GEdgar May 19 '12 at 19:19

No, it doesn't exist. In general the Laplace transform of $t^n$ is $\frac{\Gamma(n+1)}{s^{n+1}}$, and $\Gamma(n)$ isn't defined on $0,-1,-2,-3...$ This integral is the definition of the Laplace transform, so the transform doesn't exist if the integral doesn't. While there are other integral transforms that could transform $\frac{1}{t}$ in a useful way, anything other than what you gave wouldn't be considered a Laplace transform anymore.

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    $\begingroup$ Or more simply: $t^n \stackrel{ \mathcal{L}}{ \Longleftrightarrow} \frac{n!}{s^{n+1}}$ without involving the gamma ($\:\Gamma\:$) function. $\endgroup$ – night owl Jun 30 '12 at 8:10

You can actually simplify it further by substituting $st = x$, so you'll get

$$L \left(\frac 1 t \right) = \int \limits _0 ^\infty \frac {\Bbb e ^{-x}} x \ \Bbb d x$$

which is a divergent integral. In other words, the transform doesn't converge for any value of $s$.


$L\left[\frac{f(t)}{t}\right](s)=\int_s^{+\infty}f(u)du$ so $L\left[\frac{1}{t}\right](s)=\int_s^{+\infty}\frac{du}{u}=\ln u|_s^{+\infty}=\ln(s)-\ln(+\infty)$ as $\ln(+\infty)$ is not defined, then $L\left[\frac{1}{t}\right]$ is also not defined but it can be solved when $L\left[\frac{1}{t}-\frac{\cos t}{t}\right]$ or more .


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