# Laplace transform converge

1. Nov 5, 2012

### matematikuvol

$\mathcal{L}\{f(t)\}=F(s)$

$$\mathcal{L}\{e^{at}\}=\frac{1}{s-a},Re(s)>a$$
$$\mathcal{L}\{\sin (at)\}=\frac{a}{s^2+a^2}, \quad Re(s)>0$$
$$\mathcal{L}\{\cos (at)\}=\frac{s}{s^2+a^2},Re(s)>0$$

If we look at Euler identity $e^{ix}=\cos x+i\sin x$, how to get difference converge intervals $Re(s)>a$ and $Re(s)>0$?

2. Nov 5, 2012

### jasonRF

The problem is that you haven't defined things precisely enough. In this line you are assuming a is real, so the region of convergence only holds for real a. If you allow a to be a general complex number (which is required for your complex exponentials) then you get,
$$\mathcal{L}\{e^{at}\}=\frac{1}{s-a},Re(s)>Re(a).$$

If $a = i x$ with real $x$, then $Re(a)=0$ so the region of convergence is $,Re(s)>0$.

jason