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

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

wioBest ResponseYou've already chosen the best response.0
I'm guessing that this involve integration by parts...
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
do i have to reverse the order of integration ?
 one year ago

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

UnkleRhaukusBest ResponseYou've already chosen the best response.0
dw:1353224604904:dw
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
have to reverse the order of integration correctly ?
 one year ago

wioBest ResponseYou've already chosen the best response.0
is u still constant with respect to t then?
 one year ago

wioBest ResponseYou've already chosen the best response.0
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...?
 one year ago

wioBest ResponseYou've already chosen the best response.0
Is there anyway to verify the answer you're getting at?
 one year ago

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

wioBest ResponseYou've already chosen the best response.0
Okay nevermind what I said before, it looks like what you are doing is correct.
 one year ago

wioBest ResponseYou've already chosen the best response.0
What are you getting for that inner integral?
 one year ago

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

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i think i have made a mistake but im not sure where
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1353228810868:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1353228991552:dw
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
H(tu) is the Heaviside unit step function?
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1353229734181:dw
 one year ago

UnkleRhaukusBest 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!
 one year ago
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