Question

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

Answer 1

\[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\
&=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\
&=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\
&=\\
\end{align*}
\]

Answer 2

\(\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\
&=\int\limits_0^\infty e^{-pt}[\int \limits_0^tf(t-u)g(u)du]dt \\ &=\int\limits_0^\infty g(u)[\int \limits_u^\infty e^{-pt}f(t-u)dt]du \\ by \:changing \:the\:order\:of \:integration
\end{align*}\)
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....

Answer 3

@UnkleRhaukus are u trying that ^ ?

Answer 4

yeah im working on it

Answer 5

:)

Answer 6

numbers ? which numbers ?

Answer 7

\[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}&=\int\limits_0^\infty\int\limits_0^tf(t-u)g(u)\cdot\text du\cdot e^{-pt}\text dt\\
&=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\
&=\int\limits_0^\infty\int\limits_0^\infty f(t-u)g(u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\
&=\int\limits_0^\infty g(u)\int\limits_0^\infty f(t-u)H(t-u)e^{-pt}\cdot\text dt\cdot\text du\\
\text{let }v=t-u&\\
\text dv =\text dt\\
t=0\rightarrow v=-u\\
t=\infty \rightarrow v=\infty\\
&=\int\limits_0^\infty g(u)\int\limits_{-u}^\infty f(v)H(v)e^{-p(v+u)}\cdot\text dv\cdot\text du\\
&=\int\limits_0^\infty g(u)\int\limits_{-u}^\infty f(v)e^{-p(v+u)}\cdot\text dv\cdot\text du\\
\end{align*}\]

Answer 8

didn't change the order of integration.....?
plus you H is throwing me of...why(and how) its used there ... ?

Answer 9

im not sure wether or not the unsure the unit heaviside step function

Answer 10

if u follow (and understand) what i am suggesting, u don't need H

Answer 11

\[\begin{align*}\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}
&=\int\limits_0^\infty\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot e^{-pt}\cdot\text dt\\
&=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text du\cdot\text dt\\
&=\int\limits_0^\infty e^{-pt}\int\limits_0^t f(t-u)g(u)\cdot\text dt\cdot\text du\\
\text{let }v=t-u&\\
\text dv =\text dt\\
t=0\rightarrow v=-u\\
t=\infty \rightarrow v=\infty\\
&=\int\limits_0^\infty g(u)\int\limits_{u}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du\\
&=\\
\end{align*}\]

Answer 12

still, didn't change the order of integration.......

Answer 13

u=0 to u=t
cahnges to t=u to t=infinity

Answer 14

then put t-u=v

Answer 15

this question is confusing me

Answer 16

whats the confusion about ? changing order of integration ?

Answer 17

yes