Inverse Laplace Transform of (1)/(e^s-1) and (s)/(e^s-1)

ymochelows

ymochelows

Answered question

2022-09-11

Inverse Laplace Transform of 1 e s 1 and s e s 1

Answer & Explanation

Giancarlo Callahan

Giancarlo Callahan

Beginner2022-09-12Added 12 answers

It's not hard to check that
0 n = 1 δ ( t n ) e s t = n = 1 e n s = 1 e s 1
and
0 n = 1 δ ( t n ) e s t = n = 1 s e n s = s e s 1
and hence from the Laplace uniqueness theorem
L 1 [ 1 e s 1 ] ( t ) = n = 1 δ ( t n )
L 1 [ s e s 1 ] ( t ) = n = 1 δ ( t n )
which is your original train of delta functions.

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