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

Differential Equations
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\[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ &=\\ \end{align*} \]
\(\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\ &=\int\limits_0^\infty e^{-pt}[\int \limits_0^tf(t-u)g(u)du]dt \\ &=\int\limits_0^\infty g(u)[\int \limits_u^\infty e^{-pt}f(t-u)dt]du \\ by \:changing \:the\:order\:of \:integration \end{align*}\) the region from u=0 to u=t is same as region from t=u to t=\(\infty\) now, substitute t-u = x in inner integral. then you will be able to split the integrals , one only containing t, and other only containing x....
@UnkleRhaukus are u trying that ^ ?

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yeah im working on it
:)
numbers ? which numbers ?
\[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ &=\int\limits_0^\infty g(u)\int\limits_0^\infty f(t-u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\ \text{let }v=t-u&\\ \text dv =\text dt\\ t=0\rightarrow v=-u\\ t=\infty \rightarrow v=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_{-u}^\infty f(v)H(v)e^{-p(v+u)}\cdot\text dv\cdot\text du\\ &=\int\limits_0^\infty g(u)\int\limits_{-u}^\infty f(v)e^{-p(v+u)}\cdot\text dv\cdot\text du\\ \end{align*}\]
didn't change the order of integration.....? plus you H is throwing me of...why(and how) its used there ... ?
im not sure wether or not the unsure the unit heaviside step function
if u follow (and understand) what i am suggesting, u don't need H
\[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\} &=\int\limits_0^\infty\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot\text dt\\ &=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text dt\cdot\text du\\ \text{let }v=t-u&\\ \text dv =\text dt\\ t=0\rightarrow v=-u\\ t=\infty \rightarrow v=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_{u}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\\ &=\\ \end{align*}\]
still, didn't change the order of integration.......
u=0 to u=t cahnges to t=u to t=infinity
then put t-u=v
this question is confusing me
whats the confusion about ? changing order of integration ?
yes
|dw:1353245519310:dw|
\[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\} &=\int\limits_0^\infty\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot\text dt\\ u=0\rightarrow t=u\\ u=t \rightarrow t=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_u^\infty e^{-pt}f(t-u)\cdot\text dt\cdot\text du\\ \text{let }v=t-u&\\ \text dv =\text dt\\ t=0\rightarrow v=-u\\ t=\infty \rightarrow v=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_{u}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\\ \end{align*}\]
yes, now u see, how they can be separated into 2 different integrals, both going from 0 to infinity..
and both integrals are definition of LT
\[=\int\limits_0^\infty g(u)\int\limits_{u}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\] \[=\int\limits_0^\infty g(u)e^{-pu}\cdot\text du\quad\times\quad \int\limits_{u}^\infty e^{-pv}f(v)\cdot\text dv\quad?\]
=L(g(x))L(f(x)) is the most simplified form...
when u put v= t-u the limits change to 0 to infinity...
i dont understand that bit
v=t-u when t=u v= 0 when t=infty v=infty
everything else is correct.. just limits will be 0 to infinity
that is starting to make sense now
\(=\int\limits_0^\infty g(u)e^{-pu}\cdot\text du\quad\times\quad \int\limits_{0}^\infty e^{-pv}f(v)\cdot\text dv\quad =L[g(t)]L[f(t)]\)
ok ?
is this all right now? \[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\} &=\int\limits_0^\infty\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot\text dt\\ u=0\rightarrow t=u\\ u=t \rightarrow t=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_u^\infty e^{-pt}f(t-u)\cdot\text dt\cdot\text du\\ \text{let }v=t-u\\ \text dv =\text dt\\ t=u\rightarrow v=0\\ t=\infty \rightarrow v=\infty\\ &=\int\limits_0^\infty g(u)\int\limits_{0}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\\ &=\int\limits_0^\infty g(u)e^{-up}\cdot\text du\times\int\limits_{0}^\infty f(v)e^{-pv}\cdot\text dv\\ \\&=G(p)F(p) \end{align*}\]
yes
finally!
Thank you so much!
what a convoluted process

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