1. ## Laplace Transform

Hi, does anyone happen to know how to prove this equality.

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

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

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

3. 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

4. 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:

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

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

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

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

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

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

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

6. Thanks for the help.