Show that the inverse Laplace transform of F(s) = log s is given by f(x) = -1/x

Amaris Orr

Amaris Orr

Answered question

2022-09-23

Show that the inverse Laplace transform of F ( s ) = log s is given by f ( x ) = 1 x

Answer & Explanation

Voglomy8

Voglomy8

Beginner2022-09-24Added 3 answers

L 1 [ F ( s ) ] = x L 1 [ F ( s ) ] = x f ( x )
or
L 1 [ F ( s ) ] = 1 x L 1 [ F ( s ) ]
Let, F ( s ) = l o g s. So, F ( s ) = 1 s
Thus
L 1 [ F ( s ) ] = L 1 [ l o g ( s ) ] = 1 x L 1 [ 1 s ] = 1 x .1 = 1 x = f ( x )
gobeurzb

gobeurzb

Beginner2022-09-25Added 2 answers

Let f n ( x ) = 1 x 1 x > 1 / n + C n δ ( x ) where C n = 1 / n e x x d x and δ is the Dirac delta distribution satisfying L [ δ ( x ) ] ( s ) = 1. Then
lim n L [ f n ( x ) ] ( s ) = lim n 1 / n 1 x e s x d x + C n = lim n 1 / n e x e s x x d x = 0 e x e s x x d x = 0 1 s e z x d z d x = 1 s 0 e z x d x d z = 1 s 1 z d z = log s
Thus the inverse Laplace transform of log s is lim n f n ( x ) with the limit taken in the sense of distributions. Concretely it means if G ( s ) = L [ g ( x ) ] ( s ) with g integrable and C 1 then G ( s ) log s = L [ h ( x ) ] ( s ) where h ( x ) = lim n g f n ( x ) (convolution).
Also lim n f n is the distributional derivative of ( γ log x ) 1 x > 0

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