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For the purpose of my research on persistent random walks I need to compute the inverse Laplace transform of $$ F(s)=\frac{\mathrm e^{-b\sqrt{s^2-1}}}{s^2-1}.$$ I looked up in tables of integral transforms such as Erdélyi et al., Gradsteyn & Ryzhik, Prudnikov et al. with no success. So I have looked for a proper contour for the Bromwich integral. I came up with this one

Double keyhole contour

$\color{green}A$ is the integral we want to compute $$f(t)=\frac1{2\pi\mathrm i}\int_{\gamma-\mathrm i\infty}^{\gamma+\mathrm i\infty}\mathrm e^{st}\frac{\mathrm e^{-b\sqrt{s^2-1}}}{s^2-1}\mathrm ds,$$ $\color{blue}B$ and $\color{blue}K$ vanish when the radius of the large circle goes to infinity. The contributions of $\color{brown}F$ and that of $\color{brown}D$ and $\color{brown}H$ are given by simple poles and yield together a contribution of $\sinh t$. The contribution of $\color{red}C$ and $\color{red}J$ cancel each other.

To compute the contributions of $\color{red}E$ and $\color{red}G$, I set $s=-\cos\theta+\mathrm i\varepsilon$ for $\color{red}E$ and $s=\cos\theta-\mathrm i \varepsilon$ for $\color{red}G$. I end up with the following integral $$\color{red}{E+G}\to\frac1{\pi\mathrm i}\;\text{vp.}\int_0^\pi \frac{\mathrm e^{\mathrm ib\sin\theta}}{\sin\theta}\sinh(t\cos\theta)\mathrm d\theta\tag{1}$$ that I didn't manage to reduce to a handier result.

Can someone tell me if I'm the right track and perharps help me to come up with a better expression of (1) ?

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  • $\begingroup$ Maybe it helps to take the derrivative of (1) wrt. to $b$. Then you get rid of the sin in the denominator and get something that seems to go in the direction of Bessel functions. $\endgroup$
    – Urgje
    Sep 21 '15 at 11:08
  • $\begingroup$ @Urgje. Thank you for your comment. Unfortunately I already know the Laplace inverse of $s\mapsto\mathrm e^{-b\sqrt{s^2-1}}/\sqrt{s^2-1}$, it is $t\mapsto I_0(\sqrt{t^2-b^2})\Theta(t-b)$ that I can't integrate with respect to $b$ under this form. $\endgroup$
    – Tom-Tom
    Sep 21 '15 at 11:48
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I think you played a little fast and loose with the contributions over $E$ and $G$. In considering the contour integral over $E$, I let $z=e^{i \pi} x$ and, over $G$, I let $z=e^{-i \pi} x$. The integrals I get, after taking appropriate limits, is

$$e^{i \pi} \int_1^{-1} dx \frac{e^{i b \sqrt{1-x^2}}}{-(1-x^2)} e^{-x t} + e^{-i \pi} \int_{-1}^1 dx \frac{e^{-i b \sqrt{1-x^2}}}{-(1-x^2)}e^{-x t}$$

Note that the $\pm i$ in the exponentials comes from the assignment of $\sqrt{s-1} = \sqrt{e^{\pm i \pi} (x+1)}$ from the respective branch. The result is

$$-i 2 \int_{-1}^1 dx \frac{\sin{b \sqrt{1-x^2}}}{1-x^2}e^{-x t}$$

This integral converges, although I have not been able to unwind it into a closed form. (I suspect it is some Bessel of complex argument.)

Anyway, I get that

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+ i \infty} ds \frac{e^{-b \sqrt{s^2-1}}}{s^2-1} e^{s t} = \sinh{t} + \frac1{\pi} \int_{-1}^1 dx \frac{\sin{b \sqrt{1-x^2}}}{1-x^2}e^{-x t}$$

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  • $\begingroup$ Thanks a lot ! Indeed, I used a bad parametrization for $\color{red}E$, I should have used $s=\cos\theta+\mathrm i\varepsilon$ instead. Finding a closed form might be impossible, but the expression you have found will be definitely useful. $\endgroup$
    – Tom-Tom
    Sep 22 '15 at 7:50
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I have a copy of G.E.Roberts and H.Kaufman "Table of Laplace Transforms", 1966.

On page 252, Item 3.2.55 is a more complex example but might apply if $v = 0$.

Inverse of $(s + (s^2 - a^2)^{1/2})^v \exp(-b\sqrt{s^2-a^2})/\sqrt{s^2 - a^2}$ is given as:

$0$ for $0 < t < b$

and $a^v ((t-b)/(t+b))^{v/2} I_v(a\sqrt{t^2-b^2})$ for $t > b$

With the normal restrictions that $b>0$, $|\operatorname{Re}(v)| < 1$, $\operatorname{Re} s > |\operatorname{Re} a|$.

$I_v$ is the modified Bessel function of order $v$.

I assume that the expressions would be valid for $v = 0$.

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  • $\begingroup$ Thank you for your answer. I also found this inverse in the tables of Erdélyi et al (Bateman) or Prudnikov et al. But in this case, the denominator is $\sqrt{s^2-a^2}$ and not $s^2-a^2$ as in my original question (see my reply to Urgje below the question). $\endgroup$
    – Tom-Tom
    Nov 25 '15 at 8:18

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