Complex Analysis: Inverse Laplace Transform

[Niles]

Homework Statement

Hi all.

I have found the Laplace transform of the following piecewise function:

$$f(x) = \left\{ {\begin{array}{*{20}c} {0\,\,\,\,{\rm{for}}\,\,\,\,x < 0} \\ {x\,\,\,\,{\rm{for}}\,\,x \in (0;1)} \\ {0\,\,\,\,{\rm{for}}\,\,\,\,x > 1} \\ \end{array}} \right.$$

I get the Laplace transform to be

$$\overline f (s) = \frac{{ - \left( { - 1 + e^{ - s} + e^{ - s} s} \right)}}{{s^2 }}.$$

Now I wish to find the inverse Laplace transform, and this is given by the Bromwich integral:

$$f(x) = - \frac{1}{{2\pi i}}\left[ {\int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{ - e^{st} }}{{s^2 }}ds + \int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{e^{s(t - 1)} }}{{s^2 }}ds + } \int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{e^{s(t - 1)} }}{s}ds} } } \right].$$

Doing these integrals, I end up with f(x) being x for x<1 and 0 for x>1, which is not correct. I have gone through the calculations many times, and I do not believe I have made an error. Here's my reasonings:

1) For the first integral, I close the line by a semi-circle in the left half-plane, and this integral gives me -x.

2) For the second and third integral, I close the line by a semi-circle in the right half-plane for x<1, and this gives me 0, since there are no poles.

For x>1 I close the line in the left halfplane, and I get x.So, as stated earlier, I end up with f(x) being x for x<1 and 0 for x>1, and my reasoning has been 100% valid so far. Can you tell me what is wrong here?
Niles.

 

[lippyka]

Homework EquationsThe Bromwich integral:f(x) = - \frac{1}{{2\pi i}}\left[ {\int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{ - e^{st} }}{{s^2 }}ds + \int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{e^{s(t - 1)} }}{{s^2 }}ds + } \int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{e^{s(t - 1)} }}{s}ds} } } \right].The Attempt at a SolutionI believe the issue here is that you are not taking into account the fact that the Laplace transform of a piecewise function is itself piecewise. That is, the inverse Laplace transform of your Laplace transform should be a piecewise function with two parts: one for x<1 and one for x>1. For x<1, the inverse Laplace transform should be f(x) = \frac{{ - \left( { - 1 + e^{ - s} + e^{ - s} s} \right)}}{{s^2 }} = x. For x>1, the inverse Laplace transform should be f(x) = 0. Thus, the final inverse Laplace transform should be f(x) = \left\{ {\begin{array}{*{20}c} {x\,\,\,\,{\rm{for}}\,\,x \in (0;1)} \\ {0\,\,\,\,{\rm{for}}\,\,\,\,x > 1} \\\end{array}} \right.