Series Expansion for Inverse Laplace Transform of Irrational Functions

[matematikawan]

I'm trying to work something on inverse Laplace transform. I need to express a transfer function F(s) to the form
$$F(s)=\frac{s^{-1} (a_0 + a_1s^{-1} + a_2s^{-2}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }$$

I can easily do it for rational function e.g.
$$\frac{s^3+2s^2+3s+1}{s+4}= \frac{s^{-1} (1+2s^{-1}+3s^{-2}+s^{-3})}{s^{-3}+4s^{-4}}$$

for some indicial e.g
$$\frac{1}{\sqrt{s^2+s}}=\frac{1}{s}(1+s^{-1})^{-1/2}$$
Expand using the binomial theorem.

My problem is how to express irrational functions such as s^-3/2 or
$$\frac{e^{-\sqrt{s}}}{s}$$
to the form of F(s).

 

[g_edgar]

$F(s)$ is regular at infinity (in the sense of complex variables) but neither $s^{-3/2}$ nor $e^{-\sqrt{s}}/s$ is.

 

[matematikawan]

That is an interesting observation that you make. If you could please explain more what regular at infinity means. I'm do not know much about complex variables theory. That's why I want to avoid dealing with Bromwich integral.

My idea is to repeatedly use
$$L^{-1} { \frac{F(s)}{s} }=\int_0^t f(t) dt$$
for evaluating inverse transform.

From Laplace transform table I know that $s^{-3/2}$ and $e^{-2\sqrt{s}}/s$ can be inverted. That's why I expect the functions could be express as F(s)

 

[g_edgar]

OK, the function

$$F(s)=\frac{s^{-1} (a_0 + a_1s^{-1} + a_2s^{-2}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }$$

has the property that $s^kF(s)$ converges to a finite nonzero limit as $s \to \infty$ for some integer $k$ . Neither of the two functions in question has that property.

 

[gato_]

There is no standart way for those transforms. You have to proceed from the integral form:
$$\frac{1}{2\pi}PV \int_{-\infty}^{\infty}{exp(st)s^{-3/2}ds}=\frac{1}{2\pi}2\sqrt{t}\int_{0}^{\infty}{exp(z)z^{-1/2-1}dz}=2\sqrt{t}\frac{1}{2\pi}\Gamma[-1/2]=-2\sqrt{\frac{t}{\pi}}$$
(for t>0)

 

[matematikawan]

Actually I'm trying to solve pde not inverse Laplace transform per se. You know after we transform the pde wrt time and solve the transformed ode in s-domain, we usually obtain quite complicated F(s). It is this F(s) that I want to invert numerically.

But from what have you have shown, may be I can learned something from it. Just need further clarification. Why is your inverse transform for s^-3/2 has a negative sign. The entry http://www.vibrationdata.com/Laplace.htm" has no such sign. And why is your integral form different from entry 1.2 of the table. Where has the integration end limit $c+i\infty$ gone?

 

[matematikawan]

OK, the function

F(s) has the property that $s^kF(s)$ converges to a finite nonzero limit as $s \to \infty$ for some integer $k$ . Neither of the two functions in question has that property.

So the converse statement is not true for those cases. Thanks g_edgar. I have to sort out this first. May be I will change the form to

$$F(s)=\frac{s^{-1} (a_0 + a_1s^{-0.5} + a_2s^{-1}+ ... }{b_0 + b_1s^{-1} + b_2s^{-2}+ ... }$$

I will try to figure out how to handle this case and come back again if I have a problem.

 

[gato_]

There is no standart way for those transforms. You have to proceed from the integral form:
$$\frac{1}{2\pi}PV \int_{-\infty}^{\infty}{exp(st)s^{-3/2}ds}=\frac{1}{2\pi}2\sqrt{t}\int_{0}^{\infty}{exp(z)z^{-1/2-1}dz}=2\sqrt{t}\frac{1}{2\pi}\Gamma[-1/2]=-2\sqrt{\frac{t}{\pi}}$$
(for t>0)

you are right abount the factor in the limits of the integral (missed another i in the denominator ). The difference in sign is probably because i chose the wrong contour

 

[gato_]

Apparently it is easier to prove the converse, I am not sure about the inversion formula. Now,
$$\int_{0}^{\infty}t^{\nu}e^{-st}dt=\frac{1}{\nu+1}\int_{0}^{\infty}z^{\nu}e^{-z}dz=\frac{\Gamma(\nu+1)}{s^{\nu+1}}$$
The correct form of the integral I posted previously is then an alternative definition for $$\Gamma$$:
$$\frac{1}{\Gamma(\nu+1)}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{s}}{s^{\nu+1}}$$
Which I didn't kow