# Region of Convergence of Bilateral Laplace Transform

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.

• 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. Aug 17 '18 at 15:30
• 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?
– Ben
Aug 20 '18 at 8:25
• I'm saying $\mathcal B[|t|^{-1/2}] = \int_{-\infty}^\infty |t|^{-1/2} e^{-s t} dt$ converges conditionally. Aug 20 '18 at 15:09

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 $f$.
• 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 $\sigma = 0$
• 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. Aug 20 '18 at 15:21