Laplace Inverse of 1 is 1/s or Deta(t) I am confused , can anybody give me confirm answer
Thanks a lot in advance
Have you tried taking the Laplace transform of those?
The Laplace transform of f(x) is, by definition, $\displaystyle \int_0^\infty e^{-sx}f(x)dx$
So what are $\displaystyle \int_0^\infty (1)e^{-sx} dx$ and $\displaystyle \int_0^\infty (\delta(x))e^{-sx}dx$?
Surely, you understand the relation between a "Laplace transform" and its inverse? (Or, for that matter, any kind of operation and it inverse!)
Yes, $\displaystyle L(1)= \int_0^\infty e^{-st}(1)dt= \left[-\frac{1}{s}e^{-st}\right]_0^\infty= \frac{1}{s}$ so 1 is the "inverse Laplace transform" of $\displaystyle \frac{1}{s}$, not 1.
The Laplace transform of [tex]\delta(x)[\tex] is $\displaystyle L(\delta(x))= \int_0^\infty e^{-st}\delta(t)dt= e^0= 1$.
Okat, NOW what is the inverse Laplace transform of "1"?