How to inverse the Laplace Transform?

[flouran]

Hi,
I know that in order to inverse a function f(s) back to its time domain counterpart, f(t), one must use the line integral, the Bromwich Integral, but I do not know how to evaluate a line integral. Does anyone know of any practical methods of evaluating the inverse Laplace transform, could point me to some guides on how to do inverse the Laplace transform, how to calculate a line integral (especially this one), and/or personally help me themselves on this post?

Really, any help is greatly appreciated b/c I have looked everywhere for some good guides, and I can't find any helpful ones. By the way, I have taken up to BC Calculus, so I know how to integrate and differentiate.

Thanks.

 

[lurflurf]

There are several basic methods
1)Tables
One collects several known transform pairs.
The known transforms and variations can be inverted.
Several commonly used pairs require cleverness to establish without recourse to other methods
2)Bromwich Integral
A line integral
3)Numerically
This is often difficult because of instability (equivalent to exponential regression)
4)Real inversion formula
Useful for several common pairs.
Difficult to apply to harder pairs.


https://www.physicsforums.com/showthread.php?t=81279

 

[flouran]

Thank you very much. But, I am unsure of some of your steps. You said in the link you gave me that,

we can clean the integral up with a substitution i u=s t
f(t)=\frac{1}{2{\pi}\sqrt{it}}\int_{-\infty}^{\infty} \frac{e^{iu}}{\sqrt{u}} du
This integral can be written in terms of "know" real integrals.
\int_0^\infty \frac{sin(x)}{\sqrt{x}} dx=\int_0^\infty \frac{cos(x)}{\sqrt{x}} dx=\sqrt{\frac{\pi}{2}}

.

But how do you know how to write the integral in terms of "know" integrals, did you use Euler's identity by any chance to achieve this result?

 

[lurflurf]

yes actualy Euler's formula
exp(i x)=cos(x)+i sin(x)
Of course if those integrals are not "known" they must be computed

If you do that inversion with the real inversion formula
f(t)=\lim_{k\rightarrow\infty}\frac{(-1)^k}{k!}g^{(k)}(\frac{k}{t})(\frac{k}{t})^{k+1}

you will need to remember the Wallis pi formula
lim sqrt(2k+1)(2k)!/(2k+1)!=sqrt(pi/2)