Convergence of this Laplace transformation

[lukka98]

I have a f(t) that is, e^(-t) *sin(t), now I calculate the Laplace transformation, that is:
X(s) = 1 / ( 1 + ( 1 + s)^2 ) (excuse me but Latex seems not run ).
Now I imagine the plane with Re(s), Im(s) and the magnitude of X(s).

If i take Re(s) = -1 and Im(s) = 0, I believe I have X(s) = 1 ( s = -1, so from the formula X(s) = 1) and this seem correctly according to a graph that I see online.

But if I put Re(s) = -1 and Im(s) = 0 in the integral, and I calculate it directly, I should get the same result...(?) but In that case is only the integral of sin(t) from 0 to infty that is not 1 absolutely.

What is wrong?

thanks

 

[pasmith]

The integral $$\int_0^\infty \sin t\,e^{-(1+s)t}\,dt$$ converges only when $\operatorname{Re}(1 + s) > 0$. $s = -1$ is not in this region.

You can, of course, analytically continue the result you get in $\operatorname{Re}(1 + s) > 0$ into the rest of the $s$-plane (except the poles at $s = -1 \pm i$).

LaTeX is enabled; but it often doesn't render when you make your first post in a forum.

 

[lukka98]

The integral $$\int_0^\infty \sin t\,e^{-(1+s)t}\,dt$$ converges only when $\operatorname{Re}(1 + s) > 0$. $s = -1$ is not in this region.

You can, of course, analytically continue the result you get in $\operatorname{Re}(1 + s) > 0$ into the rest of the $s$-plane (except the poles at $s = -1 \pm i$).

LaTeX is enabled; but it often doesn't render when you make your first post in a forum.

Ok, maybe my question was misunderstood, but $X(s) = \frac{1}{1 + (1+s)^2}$, so for s = -1, I have X(s) = 1;
but $\int_{0}^{\infty} sin(t) dt$ ,that is for s = -1, is indefinite.
And also the graphics solution, looks like X(s) = 1 for s = -1.

 

[pasmith]

Ok, maybe my question was misunderstood, but $X(s) = \frac{1}{1 + (1+s)^2}$, so for s = -1, I have X(s) = 1;
but $\int_{0}^{\infty} sin(t) dt$ ,that is for s = -1, is indefinite.
And also the graphics solution, looks like X(s) = 1 for s = -1.

Yes.

There are many functions, such as the Gamma function, which are initially defined by an integral representation valid only in a subset of $\mathbb{C}$ but are then extended to a larger subset of $\mathbb{C}$ by analytic continuation. Laplace transforms often also fall into this category.

One can define a function $$F : D \to \mathbb{C}: s \mapsto \int_0^\infty f(t)e^{-st}\,dt$$ where $D \subset \mathbb{C}$ is the region where the integral converges. If $D$ satisfies certain conditions then there exists a unique $G$, defined on as much of $\mathbb{C}$ as possible, such that $G(s) = F(s)$ for every $s \in D$.

Now $F$ and $G$ are distinct functions because they have different domains, and we commonly define the Laplace transform of $f$ as $F$ when really we mean $G$: Inverting the transform requires integration over a contour which is almost certainly not contained in $D$, and if we have a formula for $F(s)$ then it almost certainly makes sense for values of $s \notin D$. But we must remember that if $s \notin D$ then $$G(s) \neq \int_0^\infty f(t)e^{-st}\,dt.$$