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UnkleRhaukus
 3 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
 3 years ago
Use transform definitions, and the evaluation of a suitable double integral, to calculate the Laplace transform of the following integral \[\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\}\]

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UnkleRhaukus
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0I'm guessing that this involve integration by parts...

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

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

UnkleRhaukus
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353224604904:dw

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

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

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Is there anyway to verify the answer you're getting at?

UnkleRhaukus
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0What are you getting for that inner integral?

UnkleRhaukus
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0i think i have made a mistake but im not sure where

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

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

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

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

UnkleRhaukus
 3 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!
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