1
$\begingroup$

The question is:
Compute the laplace transform of $f(t) = e^{-at}$ and state the domain of the Laplace Transform where $a\in\mathbb{C}$.

I computed the Laplace Transform of $f$ as
$$ \mathcal{L}f(z) = \int_{0}^\infty f(t)e^{-zt}dt \\ = \int_0^\infty e^{-at}e^{-zt}dt \\ = \int_0^\infty e^{-(a+z)t}dt \\ = \frac{1}{a+z}. $$
I'm not sure what information to use to find the domain of this function... is it just $z \neq -a$?

$\endgroup$
1
  • 2
    $\begingroup$ When would the integral you have diverge? $\endgroup$
    – Gregory
    Commented Oct 18, 2017 at 3:00

1 Answer 1

Reset to default
3
$\begingroup$

Note that we have

$$I(L)=\int_0^L e^{-(a+z)t}\,dt=\frac{1-e^{-(a+z)L}}{a+z}$$

If $\text{Re}(a+z)>0$, then $\lim_{L\to \infty}I(L)=\frac1{a+z}$.

If $\text{Re}(a+z)\le 0$, then $\lim_{L\to \infty}I(L)$ fails to exist.

$\endgroup$
2
  • $\begingroup$ Why do we only care about the real part of $a+z$ and not the imaginary part? $\endgroup$
    – Twenty-six colours
    Commented Oct 18, 2017 at 3:38
  • $\begingroup$ If $a+z=x+iy$, where $x\in \mathbb{R}$ and $y\in \mathbb{R}$, then we have $$I(L)=\int_0^L e^{-(a+z)t}\,dt=\frac{1-e^{-xL}e^{-iyL}}{x+iy}$$As $L\to\infty$, the exponential term in the numerator tends to $0$ if and only if $x>0$. If $x>0$ the exponential term approaches infinity and the limit fails to exist since the term $e^{iyL}$ oscillates indefinitely around the unit circle. If $x=0$, the limit fails to exist since the term $e^{iyL}$ oscillates indefinitely around the unit circle. $\endgroup$
    – Mark Viola
    Commented Oct 18, 2017 at 20:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .