## UnkleRhaukus 3 years ago Use transform definitions, and the evaluation of a suitable double integral, to calculate the Laplace transform of the following integral (ii)$\mathcal L\left\{\int\limits_0^tf(t-u)g(u)\cdot\text du\right\}$

1. UnkleRhaukus

\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*}

2. gohangoku58

\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....

3. gohangoku58

@UnkleRhaukus are u trying that ^ ?

4. UnkleRhaukus

yeah im working on it

5. gohangoku58

:)

6. gohangoku58

numbers ? which numbers ?

7. UnkleRhaukus

\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*}

8. gohangoku58

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

9. UnkleRhaukus

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

10. gohangoku58

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

11. UnkleRhaukus

\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*}

12. gohangoku58

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

13. gohangoku58

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

14. gohangoku58

then put t-u=v

15. UnkleRhaukus

this question is confusing me

16. gohangoku58

whats the confusion about ? changing order of integration ?

17. UnkleRhaukus

yes

18. gohangoku58

|dw:1353245519310:dw|

19. UnkleRhaukus

\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*}

20. gohangoku58

yes, now u see, how they can be separated into 2 different integrals, both going from 0 to infinity..

21. gohangoku58

and both integrals are definition of LT

22. UnkleRhaukus

$=\int\limits_0^\infty g(u)\int\limits_{u}^\infty e^{-p(v+u)}f(v)\cdot\text dv\cdot\text du$ $=\int\limits_0^\infty g(u)e^{-pu}\cdot\text du\quad\times\quad \int\limits_{u}^\infty e^{-pv}f(v)\cdot\text dv\quad?$

23. gohangoku58

=L(g(x))L(f(x)) is the most simplified form...

24. gohangoku58

when u put v= t-u the limits change to 0 to infinity...

25. UnkleRhaukus

i dont understand that bit

26. gohangoku58

v=t-u when t=u v= 0 when t=infty v=infty

27. gohangoku58

everything else is correct.. just limits will be 0 to infinity

28. UnkleRhaukus

that is starting to make sense now

29. gohangoku58

$$=\int\limits_0^\infty g(u)e^{-pu}\cdot\text du\quad\times\quad \int\limits_{0}^\infty e^{-pv}f(v)\cdot\text dv\quad =L[g(t)]L[f(t)]$$

30. gohangoku58

ok ?

31. UnkleRhaukus

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*}

32. gohangoku58

yes

33. gohangoku58

finally!

34. UnkleRhaukus

Thank you so much!

35. UnkleRhaukus

what a convoluted process