# Integration of Laplace transform

1. Apr 10, 2013

### Mangoes

1. The problem statement, all variables and given/known data

Find the inverse transform of the function

$$F(s) = log\frac{s-2}{s+2}$$

2. Relevant equations

$$L(\frac{f(t)}{t}) = \int^{∞}_{s}F(x)dx$$

$$f(t) = tL^{-1}(\int^{∞}_{s}F(x)dx)$$

3. The attempt at a solution

I missed the lecture on this and while I was able to figure out differentiation of transforms I've been unable to get this right. The textbook introduces the definition with the conditions necessary for the Laplace transform of f(t)/t, states the two formulas above, gives one example and then finishes the section.

$$L^{-1}(log(\frac{s-2}{s+2}))$$

$$tL^{-1}(\int^{∞}_{s}log(\frac{x-2}{x+2}) dx)$$

The main problem I'm having here is with the integrand.

$$log(x-2) - log(x+2)$$

I can easily integrate any of the two with integration by parts. Since both parts are similar, I'll just pick log(s-2).

Letting u = log(x-2) and dv = 1

$$[xlog(x-2)]^{∞}_{s} - \int^{∞}_{s}\frac{x}{x-2}$$

The amount of problems coming up by doing this is making me think I'm applying the Laplace transform wrong. If I go back now and look at the entire thing:

$$tL^{-1}(\int^{∞}_{s}\frac{-x}{x-2} dx + [xlog(x-2)]^{∞}_{s} - \int^{∞}_{s}log(x+2) dx )$$

The first term integrates into x + 2log(x-2). I have no idea how to apply the inverse Laplace tranform to a logarithm though and judging by the previous sections and problems, I'm not supposed to.

Even if I figured out how to somehow apply the inverse Laplace transform to the first term, the second term diverges when evaluating the limits of integration.

I figure I'm going at this completely wrong somewhere in the beginning, but where?

Last edited: Apr 10, 2013
2. Apr 11, 2013

### fzero

The integral for the inverse transform of a logarithm is hard to do, but the integral for $1/(s-a)$ is much easier. Try to find a formula for the inverse transform of $F'(s)$ if $F(s)$ is the Laplace transform of $f(t)$. You should be able to find the right formula by differentiating the usual integral expression for $F(s)$.