# Simple Laplace transform problem

• Nov 4th 2009, 12:33 AM
Evales
Simple Laplace transform problem
Exam study haha

2. Consider
f(t) = e^at for t > 0, where 'a' is a constant, then

F
(s) = Integral(from 0 to infinity) e^(st)e^(at)dt

and this integral exists for
s > a, so that

L
(eat) = 1/ (s-a)

for
s > a.

I would have said 1/ (a-s)
Why is it not so?

• Nov 4th 2009, 03:06 AM
mr fantastic
Quote:

Originally Posted by Evales
Exam study haha

2. Consider f(t) = e^at for t > 0, where 'a' is a constant, then

F
(s) = Integral(from 0 to infinity) e^(st)e^(at)dt

and this integral exists for

s > a, so that
L

(eat) = 1/ (s-a)

for s > a.

I would have said 1/ (a-s)
Why is it not so?

$\displaystyle \int_{0}^{+\infty} e^{-st} e^{at} \, dt = \int_{0}^{+\infty} e^{-(s - a)t} \, dt = ....$

• Nov 4th 2009, 03:15 AM
HallsofIvy
$\displaystyle \int_0^\infty e^{at}e^{-st}dt= \int_0^\infty e^{(a-s)t}dt$

Let u= (a-s)t so that du= (a-s)dt and du/(a-s)= dt. Now the integral is $\displaystyle \frac{1}{a-s}\int e^udu$ but we have to be careful about the limits of integration. At t= 0, u= (a-s)t= (a-s)0= 0 but if s> a, a-s< 0 so the upper limit will be $\displaystyle -\infty$.

That is, the integral is now $\displaystyle \frac{1}{a-s}\int_0^{-\infty} e^u du$. Integrating that gives, again, an exponential which will be 0 at the upper limit. The integral is $\displaystyle 0- \frac{1}{a-s}e^0= \frac{1}{s-a}$.