# Thread: laplace inverse

1. ## laplace inverse

Intergral from p to infinity of F(p)dp = laplace of f(t)/t
I have to reverse the order of integration to show that those two are equivalent if F(p) = laplace of f(t)

Not sure how to do this, If i reverse the order of integration is it the integral from 0 to t?

Do i use the convolution theorem where g(u) = 1?

these are the ideas I have but not sure if it is right now not

2. Originally Posted by CookieC
Intergral from p to infinity of F(p)dp = laplace of f(t)/t
I have to reverse the order of integration to show that those two are equivalent if F(p) = laplace of f(t)

Not sure how to do this, If i reverse the order of integration is it the integral from 0 to t?

Do i use the convolution theorem where g(u) = 1?

these are the ideas I have but not sure if it is right now not
Are you trying to prove that if LT[f(t)] = F(p) then $\displaystyle LT\left[\frac{f(t)}{t}\right] = \int_p^{+\infty} F(u) \, du$ ? Please post the question exactly as it is written.

3. yeah that's what i'm trying to prove sorry about the badly typed question

4. Originally Posted by mr fantastic
Are you trying to prove that if LT[f(t)] = F(p) then $\displaystyle LT\left[\frac{f(t)}{t}\right] = \int_p^{+\infty} F(u) \, du$ ? Please post the question exactly as it is written.
$\displaystyle LT\left[ \frac{f(t)}{t} \right] = \int_0^{+\infty} f(t) \frac{1}{t} e^{-pt} \, dt = \int_0^{+\infty} f(t) \left[ \int_{p}^{+\infty} e^{-ut} \, du \right] \, dt$

$\displaystyle = \int_p^{+\infty} \left[ \int_{0}^{+\infty} f(t) e^{-ut} \, dt \right] \, du = \int_p^{+\infty} F(u) \, du$.

5. Where did the 1/t go after the second equals sign?

6. Originally Posted by CookieC
Where did the 1/t go after the second equals sign?
Perhaps your question should be "Where did that second integral come from that's inside the first integral?". It could be that the two questions are related ....

7. I thought the second integral just came from the definition of the laplace transform....

8. Originally Posted by CookieC
I thought the second integral just came from the definition of the laplace transform....
You really need to think harder about the help you get given. I thought it would be obvious that I was referring to the second integral that is inside of

$\displaystyle \int_0^{+\infty} f(t) \left[ \int_{p}^{+\infty} e^{-ut} \, du \right] \, dt$.