Find the laplace transform of log[x]

[mathelord]

how do i find the laplace transform of log[x]

 

[Tide]

The Laplace transform of \log t is

\int_0^{\infty}e^{-st} \log t dt

 

[mathelord]

i know that but i tink the final answer is infinity,thats ridiculus,so i need confirmation

 

[Tide]

I don't think there is a "closed form" expression for the integral but the integral should be finite since log x integrates to x log x - x which goes to 0 as x -> 0.

 

[quantumdude]

I don't think that this Laplace transform exists. A necessary condition for the existence of the Laplace transform of f(t) is that f be continuous on 0 \leq t < \infty, but \log(t) isn't even defined at t=0.

 

[Tide]

The condition for the existence of the Laplace Transform is that function must be piecewise continuous and of exponential order. In short, it has to be integrable.

 

[quantumdude]

piecewise continuous

Right, but the interval has to include t=0, where the integrand has a vertical asymptote. Doesn't that screw things up?

 

[Tide]

Right, but the interval has to include t=0, where the integrand has an infinite discontinuity. Doesn't that screw things up?

Yes, the fact that the discontinuity occurs at t = 0 poses a problem but you can define the Laplace transform by setting the lower limit to \epsilon > 0 and passing to the limit 0.

In fact, we know that the integral

\int_{0}^{\infty}\ln x e^{-x} dx = -\gamma

is just the Euler-Mascheroni constant. We can use this result to evaluate the Laplace transform:

\int_{0}^{\infty} \ln t e^{-st} dt = \int_{0}^{\infty} \frac{-\ln s +\ln x}{s}e^{-x}dx

with the result

\int_{0}^{\infty} \ln t e^{-st} dt = - \frac {\ln s + \gamma}{s}

which wasn't as bad as I first thought it would be!