A community for students.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
 4 years ago
Use transform definitions, and the evaluation
of a suitable double integral, to calculate the
Laplace transform of the following integral
\[\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\}\]
UnkleRhaukus
 4 years ago
Use transform definitions, and the evaluation of a suitable double integral, to calculate the Laplace transform of the following integral \[\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\}\]

This Question is Closed

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{pt}\cdot\text dt\\ &=\\ &=\\ &=\end{align*}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm guessing that this involve integration by parts...

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0do i have to reverse the order of integration ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not quite sure though.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^ux(u)e^{pt}\cdot\text dt\cdot\text du\\ ?&=\int\limits_0^\infty x(u)\int\limits_0^ue^{pt}\cdot\text dt\cdot\text du\\ \end{align*}\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353224604904:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0have to reverse the order of integration correctly ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is u still constant with respect to t then?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, if we say that \(X(t)\) is the antiderivative of \(x(t)\)....then \[ X(t) = \int_0^tx(u)\cdot du \]and \(X'(t)=x(t)\) So the antiderivative is going to be dependent on \(t\) and yet the derivative is constant...?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is there anyway to verify the answer you're getting at?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0wolfram tells me the answer is \[\frac{\mathcal L\{x(t)\}}{p}=\frac1p\int\limits_0^\infty x(t)e^{pt}\text dt\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay nevermind what I said before, it looks like what you are doing is correct.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What are you getting for that inner integral?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[\begin{align*}&=\int\limits_0^\infty x(u)\int\limits_0^ue^{pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u) \frac{e^{pt}}{p}\Big_0^u\cdot\text du\\ &=\int\limits_0^\infty x(u) \frac{1e^{pu}}p\cdot\text du\\\end{align*}\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0i think i have made a mistake but im not sure where

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353228810868:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353228991552:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0H(tu) is the Heaviside unit step function?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353229734181:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty x(u)H(tu)\cdot\text du\cdot e^{pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty x(u)H(tu)e^{pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u)\int\limits_0^\infty H(tu)e^{pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u)\mathcal L \Big\{H(tu)\Big\}\cdot\text du\\ &=\int\limits_0^\infty x(u)\frac{e^{up}}{p}\cdot\text du\\ &=\frac1p\int\limits_0^\infty x(u)e^{up}\cdot\text du\\ &=\frac{X(p)}p\color{red}\checkmark \\ \end{align*}\] Thankyou mahmit2012!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.