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

1. UnkleRhaukus

\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*}

2. wio

I'm guessing that this involve integration by parts...

3. UnkleRhaukus

do i have to reverse the order of integration ?

4. wio

Not quite sure though.

5. UnkleRhaukus

@Zarkon

6. UnkleRhaukus

\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*}

7. UnkleRhaukus

@mahmit2012

8. wio

Somehow u become a t

9. UnkleRhaukus

|dw:1353224604904:dw|

10. UnkleRhaukus

have to reverse the order of integration correctly ?

11. wio

is u still constant with respect to t then?

12. UnkleRhaukus

i think so

13. wio

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...?

14. wio

Is there anyway to verify the answer you're getting at?

15. UnkleRhaukus

wolfram tells me the answer is $\frac{\mathcal L\{x(t)\}}{p}=\frac1p\int\limits_0^\infty x(t)e^{-pt}\text dt$

16. wio

Okay nevermind what I said before, it looks like what you are doing is correct.

17. wio

What are you getting for that inner integral?

18. UnkleRhaukus

\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*}

19. UnkleRhaukus

i think i have made a mistake but im not sure where

20. mahmit2012

|dw:1353228810868:dw|

21. mahmit2012

|dw:1353228991552:dw|

22. UnkleRhaukus

H(t-u) is the Heaviside unit step function?

23. mahmit2012

yes

24. mahmit2012

|dw:1353229734181:dw|

25. UnkleRhaukus

\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!