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

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?

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