asymptotics from Laplace transform

Question

Suppose I know that a non-negative random variable with density $f$ has the following Laplace transform: $$\hat{f}(s)=\int_0^{\infty}e^{-st}f(t)dt=\frac{1}{\cosh(\sqrt{2s}x)}$$ where $s>0$ and $x>0$ is a parameter. I want to find the asymptotic behavior of $f(t)$ as $t\to\infty$. The standard Tauberian theorem doesn't apply since $\hat{f}(s)$ is bounded as $s\downarrow 0$. However, if we allow $s$ to be negative, there is a singularity at $s=\frac{-\pi^2}{8x^2}$ and I feel that there should be a Tauberian-type theorem that relates the asymptotic behavior of $f(t)$ as $t\to\infty$ with the asymptotic behavior of $\hat{f}(s)$ as $s\downarrow\frac{-\pi^2}{8x^2}$ along with some exponential factor involving $\frac{-\pi^2}{8x^2}$. The context of this particular problem allows other methods to be used to compute $f$ exactly so what I'm really interested in is the Tauberian-type theorem hinted at above. References to the literature are welcome.

Answer 1

Really, you are asking for the inverse LT, which by definition is

$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{s t} \, \operatorname{sech}{a \sqrt{s}} $$

where $a=\sqrt{2} x$, and $c \gt 0$ is greater than the greatest real part of any pole of the integrand.

Normally, I would take you through an integration contour in the complex plane that would allow me to evaluate the integral via Cauchy's theorem. Nevertheless, I am going to take a different approach, one that allows me to express the integrand in terms of much a sum over simpler functions that have easy inverse LTs.

I will state the following result:

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{2 n+1} \frac1{(2 n+1)^2 + b^2} = \frac{\pi}{4 b^2} \left ( 1-\operatorname{sech}{\frac{\pi b}{2}} \right )$$

This sum may be evaluated using the residue theorem, by considering the integral

$$\oint_{C_N} dz \, \frac{\pi \, \csc{\pi z}}{(2 z+1) [(2 z+1)^2 + b^2]} $$

where $C_N$ is a square contour centered at the origin of side length $2 N+1$, where $N \in \mathbb{N}$. As $N \to \infty$, the integral goes to zero. (This is a well-known result and I will not go through the details here; for those interested, see this for example.) By the residue theorem, therefore, we may write

$$2 \sum_{n=0}^{\infty} \frac{(-1)^n}{2 n+1} \frac1{(2 n+1)^2 + b^2} = -\pi \sum_k \operatorname*{Res}_{z=z_k} \frac{\csc{\pi z}}{(2 z+1) [(2 z+1)^2 + b^2]}$$

where the $z_k$ are the zeroes of the denominator of the summand, i.e., $z_1=-1/2$, $z_{2,3} = -1/2 \pm i b/2$. The residues at these poles are simple to compute and the result follows.

Let $b=(2/\pi) a \sqrt{s}$. We may then rewrite the sum as

$$\begin{align} \operatorname{sech}{a \sqrt{s}} &= 1-\frac{16 a^2 s}{\pi^3} \sum_{n=0}^{\infty} \frac{(-1)^n}{2 n+1} \frac1{(2 n+1)^2 + \frac{4 a^2 s}{\pi^2}} \\ &= \frac{\pi}{a^2} \sum_{n=0}^{\infty} \frac{(-1)^n (2 n+1)}{s+(2 n+1)^2 \frac{\pi^2}{4 a^2}} \end{align}$$

Now we may take the ILT of the desired expression, which is merely a sum over very simple expressions. The reversal of summation and integration may be justified because, for $t \gt 0$, both sum and integral converge. The ILT is therefore

$$f(t) = \frac{\pi}{a^2} \sum_{n=0}^{\infty} (-1)^n (2 n+1) \, \exp{\left [-(2 n+1)^2 \frac{\pi^2}{4 a^2} t\right ]} $$

(The expression on the right does not converge at $t=0$ because the original integral it represents also does not converge there.)

To answer your question now, the asymptotic behavior of $f$ as $t\to\infty$ is determined by the first term in the sum, as all other terms are exponentially small for such values of $t$. Thus, we have, substituting $a=\sqrt{2} x$:

$$f(t) \sim \frac{\pi}{2 x^2} \, e^{-\frac{\pi^2}{8 x^2} t} \quad \left ( t \to \infty\right )$$

One last comment about your question: keep in mind that the LT in question has an infinite number of zeroes along the negative real axis. Thus the limit in your question doesn't make any sense.