UnkleRhaukus
  • UnkleRhaukus
Use transform definitions, and the evaluation of a suitable double integral, to calculate the Laplace transform of the following integral \[\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\}\]
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{-pt}\cdot\text dt\\ &=\\ &=\\ &=\end{align*}\]
anonymous
  • anonymous
I'm guessing that this involve integration by parts...
UnkleRhaukus
  • UnkleRhaukus
do i have to reverse the order of integration ?

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anonymous
  • anonymous
Not quite sure though.
UnkleRhaukus
  • UnkleRhaukus
@Zarkon
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^ux(u)e^{-pt}\cdot\text dt\cdot\text du\\ ?&=\int\limits_0^\infty x(u)\int\limits_0^ue^{-pt}\cdot\text dt\cdot\text du\\ \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
@mahmit2012
anonymous
  • anonymous
Somehow u become a t
UnkleRhaukus
  • UnkleRhaukus
|dw:1353224604904:dw|
UnkleRhaukus
  • UnkleRhaukus
have to reverse the order of integration correctly ?
anonymous
  • anonymous
is u still constant with respect to t then?
UnkleRhaukus
  • UnkleRhaukus
i think so
anonymous
  • anonymous
Well, 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
  • anonymous
Is there anyway to verify the answer you're getting at?
UnkleRhaukus
  • UnkleRhaukus
wolfram tells me the answer is \[\frac{\mathcal L\{x(t)\}}{p}=\frac1p\int\limits_0^\infty x(t)e^{-pt}\text dt\]
anonymous
  • anonymous
Okay nevermind what I said before, it looks like what you are doing is correct.
anonymous
  • anonymous
What are you getting for that inner integral?
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*}&=\int\limits_0^\infty x(u)\int\limits_0^ue^{-pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u) \frac{e^{-pt}}{-p}\Big|_0^u\cdot\text du\\ &=\int\limits_0^\infty x(u) \frac{1-e^{-pu}}p\cdot\text du\\\end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
i think i have made a mistake but im not sure where
anonymous
  • anonymous
|dw:1353228810868:dw|
anonymous
  • anonymous
|dw:1353228991552:dw|
UnkleRhaukus
  • UnkleRhaukus
H(t-u) is the Heaviside unit step function?
anonymous
  • anonymous
yes
anonymous
  • anonymous
|dw:1353229734181:dw|
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty x(u)H(t-u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty x(u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u)\int\limits_0^\infty H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty x(u)\mathcal L \Big\{H(t-u)\Big\}\cdot\text du\\ &=\int\limits_0^\infty x(u)\frac{e^{-up}}{p}\cdot\text du\\ &=\frac1p\int\limits_0^\infty x(u)e^{-up}\cdot\text du\\ &=\frac{X(p)}p\color{red}\checkmark \\ \end{align*}\] Thankyou mahmit2012!

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