Jun 12, 2010 #1 Charles49 Messages 87 Reaction score 0 F(s) = \frac{1}{K^s} where K is a positive real.
Jun 13, 2010 #2 Count Iblis Messages 1,858 Reaction score 8 Hint: Write this as exp[-s Log(k)] Then compare this to the Laplace integral: Integral of exp(-s t) f(t) dt So, it looks like if you take f(t) to be a function that has a very large peak around t = Log(K), you'll get the correct Laplace transform up to some normalization. Now think of making this line of reasoning more precise...
Hint: Write this as exp[-s Log(k)] Then compare this to the Laplace integral: Integral of exp(-s t) f(t) dt So, it looks like if you take f(t) to be a function that has a very large peak around t = Log(K), you'll get the correct Laplace transform up to some normalization. Now think of making this line of reasoning more precise...
Jun 13, 2010 #4 Count Iblis Messages 1,858 Reaction score 8 Charles49 said: So is it \delta(t-\log(K))? That's right!