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

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

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) ?

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})

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.

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)

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)