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Let $$ \mathcal{B}(f(s)) = \int_{-\infty}^{\infty} e^{-st} f(t) dt $$ be a bilateral laplace transform for function $ f(t) $ with $ s = \sigma - i\gamma $. I have 2 questions regarding region of convergence (ROC) of bilateral laplace transform.

  1. I found a book that discuss bilateral laplace transform. It is The Laplace Transform by D. V. Widder. In chapter VI section 2, if we take $ f(t) = |t|^{-\frac{1}{2}} $, then bilateral laplace transform of $ f(t) $ converges on the line $ \sigma = 0 $ except at the origin.
  2. If we take $ f(t) $ as a pdf of some random variable and $ F(t) = \int_{-\infty}^{t} f(x) dx $ be a cdf of the random variable, then the bilateral laplace transform of $ F(t) $ $$ \mathcal{B}\left(\int_{-\infty}^{t} f(x) dx \right) = \int_{-\infty}^{\infty} e^{-st} \left( \int_{-\infty}^{t} f(x) dx \right) dt = \frac{\mathcal{B}(f(s))}{s} $$ The question is, when I try to find its ROC, I got an empty set, which means it does not have its bilateral laplace transform.

My attempt

  1. To find its ROC, we have to specify the area of $ \sigma $ where the absolute integral exist. \begin{align} \int_{-\infty}^{\infty} \left| e^{-st} |t|^{-\frac{1}{2}} \right| dt & = \int_{-\infty}^{\infty} e^{-\sigma t} |t|^{-\frac{1}{2}} dt \\ & = \int_{-\infty}^{0} e^{-\sigma t} (-t)^{-\frac{1}{2}} dt + \int_{0}^{\infty} e^{-\sigma t} t^{-\frac{1}{2}} dt \end{align} The integral in the first term exist only for $ \sigma \leq 0 $ and the integral in the second term exist only for $ \sigma \geq 0 $. So in order for $ \int_{-\infty}^{\infty} \left| e^{-st} |t|^{-\frac{1}{2}} \right| dt $ to exist, the ROC is the intersection of ROC both of the first and second term. So we get its ROC $ \sigma = 0 $. I do not understand why the book exclude the origin.

  2. Using the same way as question 1, we have to specify $ \sigma $ where the absolute integral exist \begin{align} \int_{-\infty}^{\infty} \left| e^{-st} \left( \int_{-\infty}^{t} f(x) dx \right) \right| dt & = \int_{-\infty}^{0} \left| e^{-st} \left( \int_{-\infty}^{t} f(x) dx \right) \right| dt + \int_{0}^{\infty} \left| e^{-st} \left( \int_{-\infty}^{t} f(x) dx \right) \right| dt \\ & = \int_{-\infty}^{0} e^{-\sigma t} \left| \left( \int_{-\infty}^{t} f(x) dx \right) \right| dt + \int_{0}^{\infty} e^{-\sigma t} \left| \left( \int_{-\infty}^{t} f(x) dx \right) \right| dt \\ & = \int_{-\infty}^{0} e^{-\sigma t} \left| \left( F(t) \right) \right| dt + \int_{0}^{\infty} e^{-\sigma t} \left| \left( F(t) \right) \right| dt \\ & \leq \int_{-\infty}^{0} e^{-\sigma t} dt + \int_{0}^{\infty} e^{-\sigma t} dt \\ & = lim_{t \to -\infty} \frac{e^{-\sigma t}}{\sigma} - lim_{t \to \infty} \frac{e^{-\sigma t}}{\sigma} \end{align} In order for the absolute integral to exist, limit of the first term and the second term should be exist. Limit in the first exist for $ \sigma < 0 $ and limit in the first exist for $ \sigma > 0 $. So the ROC is an empty set, or in other word, the bilateral laplace transform is not exist. But it should be equal to $ \frac{\mathcal{B}(f(s))}{s} $, which means its ROC should not be an empty set.

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  • $\begingroup$ Note that $\int_0^\infty t^{-1/2} e^{-\sigma \hspace{1px} t} dt$ converges for $\sigma > 0$, not for $\sigma \geq 0$. $\mathcal B$ converges conditionally but not absolutely. $\endgroup$
    – Maxim
    Aug 17, 2018 at 15:30
  • $\begingroup$ Oh sorry my mistake, yes after I checked again, it converges for $ \sigma > 0 $, but why is it converges conditionally? I already solve $ \int_{-\infty}^{\infty} e^{-st} | t |^{-1/2} dt $ by involving gamma function, yes it is converges. But why is integration on the absolute one does not converge? When we take the absolute one, it only change variable $ s $ with its real part, right? $\endgroup$
    – Ben
    Aug 20, 2018 at 8:25
  • $\begingroup$ I'm saying $\mathcal B[|t|^{-1/2}] = \int_{-\infty}^\infty |t|^{-1/2} e^{-s t} dt$ converges conditionally. $\endgroup$
    – Maxim
    Aug 20, 2018 at 15:09

1 Answer 1

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The bilateral transform of a pdf $f$ exists at least on the line $\sigma = 0$, but the transform of the corresponding cdf $F$ may not exist. $\mathcal B[F]$ exists for $s \in ROC(f) \land \sigma> 0$. In order for the intersection of $ROC(f)$ with $\sigma > 0$ to be non-empty, $f$ needs to decay (or, for general $f$, oscillate) rapidly at $-\infty$.

As an example, take $f(t) = 1/(\pi(1 + t^2)), \;F(t) = 1/2 + 1/\pi \arctan t$.

If $f(t) = |t|^{-1/2}$, then the integral $\mathcal B[f]$ converges only for $\sigma = 0$. You also get a singularity at $\gamma = 0$, because $\gamma$ gives the oscillatory component due to which $\mathcal B[f]$ converges (by Dirichlet's test), and if $s = 0$, $\mathcal B[f]$ diverges. But this is an integrable singularity, and the Bromwich integral over the vertical line $\sigma = 0$ converges to $f$.

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  • $\begingroup$ In my question comment, you said that the integral converges for $ \sigma > 0 $ and I agree to that, but it also makes me confuse, why the book said that it only converges for $ \sigma = 0 $, but in the other hand, the right part of the integration does not converge for $ \sigma = 0 $ $\endgroup$
    – Ben
    Aug 20, 2018 at 8:31
  • $\begingroup$ For which complex values of $s$ does $\int_0^\infty |t|^{-1/2} e^{-s t} dt$ converge? For which complex values of $s$ does the same integral converge absolutely? Then consider that we need the integral over $(-\infty, 0]$ as well. $\endgroup$
    – Maxim
    Aug 20, 2018 at 15:21

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