# Laplace Transform

• Jun 6th 2009, 05:21 PM
nonsingular
Laplace Transform
Hi, does anyone happen to know how to prove this equality.

Let $\displaystyle \displaystyle F(s) = \mathcal{L}\{f(t)\}$, then $\displaystyle \displaystyle\mathcal{L}\left\{\int_0^{\infty}\fra c {t^u \cdot f(u)}{\Gamma (u + 1)}du\right\} = \frac {F(\ln s)}{s\ln s}$
• Jun 6th 2009, 05:44 PM
NonCommAlg
Quote:

Originally Posted by nonsingular
Hi, does anyone happen to know how to prove this equality.

Let $\displaystyle \displaystyle F(s) = \mathcal{L}\{f(t)\}$, then $\displaystyle \displaystyle\mathcal{L}\left\{\int_0^{\infty}\fra c {t^u \cdot f(u)}{\Gamma (u + 1)}du\right\} = \frac {F(\ln s)}{s\ln s}$

why do i get $\displaystyle \frac{F(\ln s)}{s}$ as the answer then? (Thinking)
• Jun 6th 2009, 06:09 PM
nonsingular
I checked again to see that I did not type the problem incorrectly. Also I looked up Schaum's Mathematical Handbook of Formulas and Tables by Murray Spiegel, and on page 182, the Laplace Transform 33.23 concurred with my original problem. Hence, may you please post your solution to see how you got your answer? Thanks.

An abridged version of Spiegel's book can be found here Schaum's Outline of Mathematical ... - Google Book Search
• Jun 6th 2009, 06:42 PM
NonCommAlg
Quote:

Originally Posted by nonsingular
I checked again to see that I did not type the problem incorrectly. Also I looked up Schaum's Mathematical Handbook of Formulas and Tables by Murray Spiegel, and on page 182, the Laplace Transform 33.23 concurred with my original problem. Hence, may you please post your solution to see how you got your answer? Thanks.

An abridged version of Spiegel's book can be found here Schaum's Outline of Mathematical ... - Google Book Search

it's just a simple change of the order of integration:

$\displaystyle \mathcal{L}\left\{\int_0^{\infty}\frac {t^u f(u)}{\Gamma (u + 1)}du \right \} =\int_0^{\infty}e^{-st} \int_0^{\infty}\frac {t^u f(u)}{\Gamma (u + 1)}du \ dt=\int_0^{\infty} \frac{f(u)}{\Gamma(u+1)} \int_0^{\infty}e^{-st}t^u dt \ du$

$\displaystyle =\int_0^{\infty} \frac{f(u)}{\Gamma(u+1)} \mathcal{L}(t^u) \ du=\int_0^{\infty} \frac{f(u)}{\Gamma(u+1)} \cdot \frac{\Gamma(u+1)}{s^{u+1}} \ du=\int_0^{\infty}\frac{f(u)}{s^{u+1}} \ du$

$\displaystyle =\frac{1}{s} \int_0^{\infty}f(u)e^{- u\ln s} \ du=\frac{F(\ln s)}{s}.$
• Jun 6th 2009, 07:19 PM
mr fantastic
Quote:

Originally Posted by NonCommAlg
it's just a simple change of the order of integration:

$\displaystyle \mathcal{L}\left\{\int_0^{\infty}\frac {t^u f(u)}{\Gamma (u + 1)}du \right \} =\int_0^{\infty}e^{-st} \int_0^{\infty}\frac {t^u f(u)}{\Gamma (u + 1)}du \ dt=\int_0^{\infty} \frac{f(u)}{\Gamma(u+1)} \int_0^{\infty}e^{-st}t^u dt \ du$

$\displaystyle =\int_0^{\infty} \frac{f(u)}{\Gamma(u+1)} \mathcal{L}(t^u) \ du=\int_0^{\infty} \frac{f(u)}{\Gamma(u+1)} \cdot \frac{\Gamma(u+1)}{s^{u+1}} \ du=\int_0^{\infty}\frac{f(u)}{s^{u+1}} \ du$

$\displaystyle =\frac{1}{s} \int_0^{\infty}f(u)e^{- u\ln s} \ du=\frac{F(\ln s)}{s}.$

Not that you need a humble ant like me confirming your solution, but when I originally saw this question that's exactly what I was going to post but didn't have the time.

@OP: I really like the Schaum's Outline Series but this wouldn't be the first time I've come across a typo (my edition of the Laplace Transforms book in this series has hundreds).
• Jun 7th 2009, 03:47 AM
nonsingular
Thanks for the help.