Charles49 Messages 87 Reaction score 0 Thread starter Jun 12, 2010 #1 [tex]F(s) = \frac{1}{K^s}[/tex] where K is a positive real.
Count Iblis Messages 1,859 Reaction score 8 Jun 13, 2010 #2 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...
Count Iblis Messages 1,859 Reaction score 8 Jun 13, 2010 #4 Charles49 said: So is it [tex]\delta(t-\log(K))[/tex]? That's right!