laplace inverse

**CookieC** · October 8th 2010, 06:41 PM

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

**mr fantastic** · October 8th 2010, 10:02 PM

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.

**CookieC** · October 9th 2010, 02:33 AM

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

**mr fantastic** · October 9th 2010, 03:53 AM

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.

$\displaystyle LT\left[ \frac{f(t)}{t} \right] = \int_0^{+\infty} f(t) \frac{1}{t} e^{-pt} \, dt = \int_0^{+\infty} f(t) \left[ \int_{p}^{+\infty} e^{-ut} \, du \right] \, dt$

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

**CookieC** · October 9th 2010, 05:59 PM

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

**mr fantastic** · October 9th 2010, 07:05 PM

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 ....

**CookieC** · October 9th 2010, 07:24 PM

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

**mr fantastic** · October 9th 2010, 09:15 PM

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

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

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