- #1

Hoplite

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## Homework Statement

I want to invert a function from Laplace transform space to normal space.

## Homework Equations

In Laplace transform space, the function takes the form $$ \bar f (s) = \frac{\exp\left[ x (-a +\sqrt{a^2+ b +c s} )\right]}{-a +\sqrt{a^2+ b +c s}}.

$$

Here, ##s## is the Laplace transform space parameter ##(t\rightarrow s)##, and ##a##, ##b##, ##c##, and ##x## are all positive constants.

## The Attempt at a Solution

Naturally, my first method for inverting a Laplace transform space function is to ask Mathematica, but it can't solve this one. My second method is to break the function into to pieces and use the inverse Laplace transform property:

$$ \bar h (s) \bar g(s) \mapsto \int_0^t h(\tau ) g(t-\tau ) d\tau .

$$

However, while Mathematica can find the Laplace Transform of the numerator by itself

$$ \exp\left[ x (-a +\sqrt{a^2+ b +c s} )\right] \mapsto \frac{x e^{-\frac{t \left(a^2+b\right)}{c}+a x-\frac{c x^2}{4 t}}}{2 \sqrt{\pi } c \sqrt{\frac{t^3}{c^3}}} ,

$$

the other part of the function, it can't do:

$$ \frac{1}{-a +\sqrt{a^2+ b +c s}} \mapsto ?? .$$

(It does give an inverse Laplace transform for this function, but only for the case where ##a<0##, and I need ##a>0##.)

I also tried to solve this by finding the residues. I figure there's only one poll (at ##s =-b/c##). And Mathematica gives the residue of

$$ \frac{\exp[s t]}{-a +\sqrt{a^2+ b +c s}} $$

at ##s=-b/c## as ##0##. Could there by any other poles for this function?

I also tried calculating the residue myself by taking a series expansion of

$$

\frac{\exp[s t](s+\frac{b}{c})}{-a +\sqrt{a^2+ b +c s}}

$$

around ##s=-b/c##. But there is no constant term in that series, which I understand mean that the residue is indeed zero.

Does anyone have any suggestions for how I could inverse Laplace transform this function?