# Laplace Transform $s$ sufficiently large [duplicate]

Possible Duplicate:
How does partial fraction decomposition avoid division by zero?

I've got a question about the domain restriction of $s$ when taking the Laplace Transform. I think my question can be best illustrated with an example.

Let's suppose we have $$\frac{d^2x}{dt} + \frac{dx}{dt} = e^{3t} + e^{7t}\\ \mathcal{L}\left\{\frac{d^2x}{dt} + \frac{dx}{dt}\right\} = \mathcal{L}\{e^{3t} + e^{7t}\}$$ Assuming the initial conditions are zero. $$s^2Y(s) + sY(s) = \frac{1}{s-3} + \frac{1}{s-7} \,\forall\, s > 7\\ Y(s)s(s + 1) = \frac{1}{s-3} + \frac{1}{s-7}\\ Y(s) = \frac{1}{s(s-3)(s+1)} + \frac{1}{s(s-7)(s+1)}\\ Y(s) = \frac{A}{s} + \frac{B}{s-3} + \frac{C}{s+1} + \frac{D}{s} + \frac{E}{s-7} + \frac{F}{s+1}$$ The reason why I wrote that $s > 7$ is because $s$ must be sufficiently large for the Laplace Transform to converge, and in the case, $s > 7$ to ensure that $e^{7t}$ does infact converge as $t \to \infty$.

Now, let's say that I want to find $A$, $B$ and $C$.

So I'd say $$1 \equiv A(s-3)(s+1) + Bs(s+1) + Cs(s-3)$$ Now to find $A$, I'd have to let $s \to 0$; however, above I said $s > 7$ (otherwise the integral wouldn't converge and we wouldn't even be able to get to this stage).

How does this work? Is it even possible to find the partial fraction coefficients given that $s > 7$?

Thank you.

Edit:

This would make sense to me: $$\frac{d^2x}{dt} + \frac{dx}{dt} = e^{3t} + e^{7t}\\ \mathcal{L}\left\{\frac{d^2x}{dt} + \frac{dx}{dt}\right\} = \mathcal{L}\{e^{3t} + e^{7t}\}$$ Assuming the initial conditions are zero. $$s_1^2Y(s) + s_1Y(s) = \frac{1}{s_2-3} + \frac{1}{s_3-7}$$ Where $s_1 > 0$, $s_2 > 3$ and $s_3 > 7$.

$$Y(s)s_1(s_1 + 1) = \frac{1}{s_2-3} + \frac{1}{s_3-7}\\ Y(s) = \frac{1}{s_1(s_2-3)(s_1+1)} + \frac{1}{s_1(s_3-7)(s_1+1)}\\ Y(s) = \frac{A}{s_1} + \frac{B}{s_2-3} + \frac{C}{s_1+1} + \frac{D}{s_1} + \frac{E}{s_3-7} + \frac{F}{s_1+1}\\ \therefore 1 \equiv A(s_2-3)(s_1+1) + Bs_1(s_1+1) + Cs_1(s_2-3)$$ And now, to find $A$, I can let $s_1 \to 0^+$ because $s_1$ is defined for all numbers greater than zero.

This would make sense to me, but I'm not sure if it correct. It seems to me that when we do the Laplace Transform and let $s \to$ some undefined value, it is almost a bit of a fluke that it works out.

-

## marked as duplicate by Pedro Tamaroff, Jennifer Dylan, William, rschwieb, ArkamisSep 22 '12 at 15:47

Related. In fact, I think the answers there are valid here, too. –  Pedro Tamaroff Jul 3 '12 at 1:30

This sort of technicality gets brushed aside in lectures, but can lead to confusion.

The function $s \mapsto Y(s)$ is defined and analytic for $\mathcal{Re}(s) >7$, but is equal to the rational function given above which is analytic everywhere except at its poles. You can factor/expand the rational function any way you want (algebraically) and it will remain the same function, which will still equal $Y$ when $\mathcal{Re}(s) >7$.

There are three quantities involved above; First (with slight abuse of notation) is $Y(s)$, which is defined for $\mathcal{Re}(s) >7$, second is the rational function $q_1(s) = \frac{1}{s(s-3)(s+1)} + \frac{1}{s(s-7)(s+1)}$, and third is the rational function $q_2(s) = \frac{A}{s} + \frac{B}{s-3} + \frac{C}{s+1} + \frac{D}{s} + \frac{E}{s-7} + \frac{F}{s+1}$.

Direct computation of the Laplace Transforms shows that $Y(s) = q_1(s)$, whenever $\mathcal{Re}(s) >7$.

The functions $q_1, q_2$ are defined on $\Delta = \mathbb{C}\setminus \{-1, 0, 3, 7 \}$ (note that if $\mathcal{Re}(s) >7$, then $s \in \Delta$), and you can establish (using any argument you like, such as letting $s\to 0$) that $q_1(s) = q_2(s)$, whenever $s \in \Delta$. Note that this equality is true regardless of how you ended up with $q_1$, and that the identify is true regardless of other conditions such as $\mathcal{Re}(s) >7$.

It therefore follows that $Y(s) = q_2(s)$, whenever $\mathcal{Re}(s) >7$.

-
I sort of understand what you're saying; however, it really doesn't make sense to me that $s > 7$, yet later on, I let $s \to 0$. I would've thought that the value as $s \to 0$ would just return some undefined behavior because the function is not valid there... –  user968243 Jul 3 '12 at 3:34
I can't work out how to edit the above comment... Anyway, I've added to the question how I believe it works. It'd be great if you could shed some light on this. Thank you. –  user968243 Jul 3 '12 at 3:48
@user968243 Consider visiting the link I provide in the comments. Bill gives a neat answer. –  Pedro Tamaroff Jul 3 '12 at 3:48
I looked at the link that you posted. I thought that when you try to find the partial fraction coefficients you equate the numerators and take the limit as the numerator approaches the poles. If I understand, what copper.hat and Bill are basically saying is that we can let $s$ approach some value for which it is undefined and then we can solve for constants. Then we can use those constants in formulating an equivalent equation (despite those constants being obtained using undefined values of $s$). And basically, this has all been proven to work and is a property of polynomials and algebra. –  user968243 Jul 3 '12 at 4:19
@user968243: I have added further elaboration, I hope it works for you. –  copper.hat Jul 3 '12 at 6:12