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

Differential Equations
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\[\begin{align*}\mathcal L\left\{\int\limits_0^tx(u)\cdot\text du\right\} &=\int\limits_0^\infty\left(\int\limits_0^tx(u)\cdot\text du\right)e^{-pt}\cdot\text dt\\ &=\\ &=\\ &=\end{align*}\]
I'm guessing that this involve integration by parts...
do i have to reverse the order of integration ?

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

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