Laplace Transform of Integrals

  • Thread starter lamdc
  • Start date
  • #1
1
0
So I know the laplace transform of an integral is the product of the laplace transorm of the functions inside the integral, such as Laplace{integral from 0 to t [tau*e^(t - tau) d tau = Laplace[t] * Laplace[e^t], ignoring the tau

what if I have Laplace{integral from 0 to t [t*e^(-tau) d tau], how do I break it down into 2 laplace functions? The answer is (1/s) * (1/(s + 1)^2), I don't get it. :/


Sorry, I don't know how to type an equation but hopefully you know what I mean. Any help is appreciated.
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Do you mean
$$
\mathcal L\left(\int_0^t \tau e^{-\tau} d\tau\right)?
$$
In that case you would write this as
$$
\mathcal L \left( \int_0^t g(t-\tau) f(\tau) d\tau\right),
$$
where ##g(t-\tau) = 1## is the constant function 1 and ##f(\tau)## is the original integrand. By the convolution result you mentioned in the beginning, it follows that
$$
\mathcal L\left(\int_0^t \tau e^{-\tau} d\tau\right) = \mathcal L(g) \mathcal L(f) = \mathcal L(1) \mathcal L(t e^{-t})= \frac{1}{s} \frac{1}{(s+1)^2},
$$
which is the quoted result.

Note that this can also be shown in a different way for any integrand ##f(\tau)##, namely:
$$
\mathcal L\left( \int_0^t f(\tau) d\tau\right)
=
\int_{t = 0}^\infty \int_{\tau = 0}^t e^{-st} f(\tau) d\tau \, dt.
$$
Changing the order of the integrals results in
$$
\int_{\tau = 0}^\infty f(\tau) \int_{t = \tau}^\infty e^{-st} dt\, d\tau
= \frac{1}{s} \int_{\tau = 0}^\infty f(\tau)\left[-e^{-st}\right]_\tau^\infty d\tau
= \frac{1}{s} \int_{\tau = 0}^\infty f(\tau) e^{-s\tau} d\tau,
$$
which is equal to ##\mathcal L(f)/s## by definition.
 

Related Threads on Laplace Transform of Integrals

  • Last Post
Replies
5
Views
534
Replies
4
Views
2K
Replies
3
Views
932
Replies
2
Views
3K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
20K
  • Last Post
Replies
8
Views
2K
Top