# Laplace Transform inversion

## Main Question or Discussion Point

Hi,

I want to inverse this laplace transform, -(s^(1/2)), seems that the inverse is in complex plane. Where should i start to find this inverse....

Thank you.

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HallsofIvy
Homework Helper
According to this table, http://www.vibrationdata.com/Laplace.htm, $1/s^k$, where k can be any real number is the Laplace transform of $t^{k-1}/\Gamma(k)$ so the "inverse Laplace tranform" of $-1/s^{1/2}$ is $-t^{-1/2}/\Gamma(1/2)= -1/\sqrt{\pi t}$.

According to this table, http://www.vibrationdata.com/Laplace.htm, $1/s^k$, where k can be any real number is the Laplace transform of $t^{k-1}/\Gamma(k)$ so the "inverse Laplace tranform" of $-1/s^{1/2}$ is $-t^{-1/2}/\Gamma(1/2)= -1/\sqrt{\pi t}$.
Thank you for your reply. I'm sorry I should use latex, I'm looking inversion of $$-\sqrt{s}$$

In the first place does the inversion of $-\sqrt{s}$ exist????

Let say it exist :surprised

From the http://www.vibrationdata.com/Laplace.htm" [Broken]formula,
$$L\{ \frac{df}{dt} \} = sF(s) - f(0^-)$$

Let $f(0^-)=0$ :surprised and choose F(s)=-s-1/2 so that $f(t)=-1/\sqrt{\pi t}$

Hence
$$L^{-1} \{ -\sqrt{s} \} = \frac{d}{dt} (-1/\sqrt{\pi t}) = \frac{\pi}{2} (\pi t)^{-3/2}$$

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Mute
Homework Helper
In the first place does the inversion of $-\sqrt{s}$ exist????

Let say it exist :surprised

From the http://www.vibrationdata.com/Laplace.htm" [Broken]formula,
$$L\{ \frac{df}{dt} \} = sF(s) - f(0^-)$$

Let $f(0^-)=0$ :surprised and choose F(s)=-s-1/2 so that $f(t)=-1/\sqrt{\pi t}$

Hence
$$L^{-1} \{ -\sqrt{s} \} = \frac{d}{dt} (-1/\sqrt{\pi t}) = \frac{\pi}{2} (\pi t)^{-3/2}$$

If you choose f(t) to be t^{-1/2}, f(0-) is clearly not zero! You can't use the first derivative rule as you did because

$$\mathcal L\left(\frac{d t^{-1/2}}{dt} = -\frac{1}{2}t^{-3/2}\right)$$

doesn't exist! The Laplace transform

$$\int_{0^-}^\infty dt~t^q e^{-st}$$

exists only for $\mbox{Re}(q) > -1$, so saying

$$\int_{0^-}^\infty dt~t^{-3/2} e^{-st} \propto \mathcal L[t^{-1/2}] + f(0^-)$$

is somewhat nonsensical (though you can argue that the LHS doesn't exist because the integral diverges, which is indicated by f(0-) being undefined).

So, the transform of t^{-1/2} does exist, though you won't be able to see that from the first derivative rule.

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Hi guys.

Thank you for your reply. Actually I found one journal 'Multi-precision Laplace transform inversion' discussing about laplace inversion in complex plane named fixed talbot algorithm.

this is the direct link to the said journal. http://www.pe.tamu.edu/valko/public_html/CV/ValkoPDF/2004AV_IJNME_Multi.pdf

Talbot pioneered the approach of deforming the standard contour in the Bromwich integral

$$f(t) = \frac{1}{2\pi i} \int_B \left exp(ts) \hat{f}(s) \right ds$$

$$B$$ in above equation is a vertical line defined by $$s=r+iy$$. By Cauchy's theorem the deformed contour is valid where, line to a contour that ends in the left half plane (the integration from -infinity to infinity).

by using this method ,the laplace inversion for

$$- \sqrt{s}$$ is $$\frac{1}{2 \sqrt{\pi t^3}}$$

i still couldn't prove it manually, because couldn't understand a few concept involved. i need to discuss it with my 'sensei'

I know my argument has a lot flaw. But I did get the right answer . I was hoping someone could improve on the method.

The Laplace transform

$$\int_{0^-}^\infty dt~t^q e^{-st}$$

exists only for $\mbox{Re}(q) > -1$, so saying

$$\int_{0^-}^\infty dt~t^{-3/2} e^{-st} \propto \mathcal L[t^{-1/2}] + f(0^-)$$

is somewhat nonsensical (though you can argue that the LHS doesn't exist because the integral diverges, which is indicated by f(0-) being undefined).

So, the transform of t^{-1/2} does exist, though you won't be able to see that from the first derivative rule.
This is a mystery because when I asked a friend to get the Laplace transform of $t^{-3/2}$ using Mathematica, it is possible to do it.
$$L \{ \frac{1}{t^{3/2}} \} = -2\sqrt{\pi s}$$

One more question Mute. When you write the Laplace transform as

$$\int_{0^-}^\infty dt~t^q e^{-st}$$

Is it not that t are positive. But why do we have the limit 0-. This also appears in the first derivative rule.

Mute
Homework Helper
This is a mystery because when I asked a friend to get the Laplace transform of $t^{-3/2}$ using Mathematica, it is possible to do it.
$$L \{ \frac{1}{t^{3/2}} \} = -2\sqrt{\pi s}$$
I'm not sure. I would guess Mathematica is just giving that answer as a purely formal result based on the general formula in my last post. If you ask Mathematica to do the integral

$$\int_0^\infty dt~\frac{e^{-st}}{t^{3/2}},$$
it will return the result "Integral does not converge".

One more question Mute. When you write the Laplace transform as

$$\int_{0^-}^\infty dt~t^q e^{-st}$$

Is it not that t are positive. But why do we have the limit 0-. This also appears in the first derivative rule.
Approaching zero from the negative side is just so the laplace transform of the delta function gives 1 instead of zero. I don't think it makes a difference for other functions.

Thanks for that info about the usage of 0-.
I don't have access to Mathematica to experiment further. But if I'm to use http://integrals.wolfram.com/index.jsp?expr=Exp[-s*x]/x^(3/2)&random=false" for indefinite integral, it look like Mathematical can integrates it.

$$\int dt~\frac{e^{-st}}{t^{3/2}} = -2\sqrt{\pi s} \mbox{ erf}(\sqrt{st}) - \frac{2e^{-st}}{\sqrt{t}}$$

It still looks like the function fail to exist when t tends to zero.

Could it possible that Mathematica does not use Riemann integration? Lebesgue integration!! what's this?

And where is https://www.physicsforums.com/showthread.php?t=355612"? I know he is an expert with that complex inversion formula. Can it be done with Bromwich integral ?

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