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alamleh · Apr 25, 2008, 07:33 AM

Need to find laplace transform for
L{exp(2t)/t}

galactus · Apr 25, 2008, 08:57 AM

$-ln(s-2)=ln(\frac{1}{s-2})$

ebaines · Apr 25, 2008, 10:56 AM

Galactus - you seem to be talking about $L(e^{2t})$ , but what about the t in the denominator? How do you do $L(\frac {e^{2t}} t)$ ?

galactus · Apr 25, 2008, 11:59 AM

I included that.

$L(\frac{1}{t})=ln(\frac{1}{s})=-ln(s)$

$L(e^{2t})=\frac{1}{s-2}$

$L(\frac{e^{2t}}{t})=ln(\frac{1}{s-2})$

ebaines · Apr 25, 2008, 02:49 PM

I'm afraid I'm losing you at L(1/t) = ln (1/s). I tried calculating L(1/t) and I get infinity:

Starting with the definition L[f(t)]= F(s), and the fact that

$ L [ \frac {f(t)} t]) = \int _s ^ \infty F(s) ds \\ $

then with f(t) = 1:

$ L( \frac 1 t ) = \int _s ^ \infty (1/s) ds \ = \ ln(\infty ) - ln(s) = \infty $

Where am I going wrong?

I must admit that I last struggled with Laplace transforms 25 years ago, so I am quite rusty on this stuff.

galactus · Apr 25, 2008, 03:32 PM

I must also admit that I looked these up in a LaPlace table or ran them through my TI-92, which has a Laplace program. I didn't do it from scratch.

The Laplace form is $\int_{0}^{\infty}e^{-st}f(t)dt$

This one would be $\int_{0}^{\infty}\frac{e^{-st}e^{2t}}{t}dt$

Which is a booger.

The inside cover of a DE book I have has a bunch of them.

I just used $L(\frac{e^{at}}{t}) = -ln(s-a)$

alamleh · Apr 26, 2008, 08:56 AM

thanks all for the help,
I think I can prove it now :

L{t . 1/t} = - dF/dS where F = L { 1/t }

the left hand side is L { 1 } = 1/S

therefore, dF/dS = - 1/S which means that F= - Ln(S)

i.e. L { 1/t } = - Ln (S)

and L { exp( a.t). /t } = - Ln(S-a)