The Hardy–Littlewood Tauberian theorem for Laplace transform in Chapter XIII in "An Introduction to Probability Theory and Its Applications" by Feller reads as follows

Let $F : [0,\infty) \to \mathbb{R}$ of bounded variation, $p \geq 0$ be real number and $$\omega_F(s) = \int^\infty_0 e^{-st} d F(t).$$ Then each of the relations $$ \dfrac{\omega_F(\tau \lambda)}{\omega_F(\tau)} \to \lambda^{-p}\hspace{15pt} \text{as $\tau \to 0$}.$$ $$ \dfrac{F(tx)}{F(t)} \to x^{p} \hspace{15pt} \text{as $t \to \infty$}.$$ implies the other as well as $$ \omega_F(1/t) \sim F(t) \Gamma(p+1) \hspace{15pt} \text{as $t \to \infty$}.$$

I have three questions.

  1. First, generally, what is the condition for the existence of an inverse Laplace transform?
  2. Second, I am so doubious that this Tauberian theorem is true for $p=0$. The inverse Laplace transform of $1$ is $\delta(t)$: then, in this case is it true that $$ \dfrac{\omega_F(\tau \lambda)}{\omega_F(\tau)} \to 1 \implies \dfrac{F(tx)}{F(t)} \to H(t) $$ where $H(t)$ is the Heaviside step function, rather than converging to $1$?
  3. Finally, the third question is: do the Tauberian theorem for the Laplace transform holds in the form $$ F(s) = \displaystyle\int^\infty_0 e^{-st} f(t) dt, $$ namely does it implies the asymptotic relation between $F$ and $f$?


  • $\begingroup$ I can not find this book, could you provide the link of this? Thanks! $\endgroup$
    – mnmn1993
    Commented Jul 25, 2022 at 2:00
  • $\begingroup$ Actually, I have read several books about Laplace Transform, but I can not find the answers. $\endgroup$
    – mnmn1993
    Commented Jul 25, 2022 at 2:01
  • 2
    $\begingroup$ Sorry, my memory failed. The author is actually Widder. $\endgroup$ Commented Jul 25, 2022 at 6:42
  • $\begingroup$ Thank you so much for the help! This book settled question 2. But I can not find answers for questions 1 and 3. For question 3, I have read through chapter 5 about the tauberian theorem but I do not find out any useful stuff. For question 1, should the condition be the one tgat the mellin inverse integral is convergent? $\endgroup$
    – mnmn1993
    Commented Jul 26, 2022 at 7:54

1 Answer 1


Just to get a few points :) , let me repeat my comment : part of the answer is contained in Widder's book Laplace transform.

  • $\begingroup$ I finally get the results from the book, thanks! $\endgroup$
    – mnmn1993
    Commented Jul 31, 2022 at 14:06
  • 4
    $\begingroup$ Can you perhaps elaborate? (As it stands, this answer is not much more than a link-only answer.) $\endgroup$
    – Stefan Kohl
    Commented Jul 31, 2022 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.