What is the Laplace transform of int_0^t (sin x)/(x) dx

Cale Terrell

Cale Terrell

Answered question

2022-10-30

What is the Laplace transform of
0 t sin x x d x
I'm thinking about approaching it as a convolution but I am not sure how. Could I define it as the convolution of 1 and sin x x ?

Answer & Explanation

driogairea1

driogairea1

Beginner2022-10-31Added 16 answers

We need
F ( s ) = 0 e s t 0 t sin ( x ) x d x d t = 0 x e s t sin ( x ) x d t d x = 0 sin ( x ) x e s x s d x
We have
F ( s ) = 0 sin ( x ) x x e s x s d x 0 sin ( x ) x e s x s 2 d x
This gives us
s F + F = 0 e s x sin ( x ) d x = 1 s 2 + 1 d d s ( s F ( s ) ) = 1 s 2 + 1
s F ( s ) lim y 0 y F ( y ) = 0 s d t t 2 + 1 = arctan ( s )
We have
lim y 0 y F ( y ) = lim y 0 0 sin ( x ) x e y x d x = 0 lim y 0 e y x sin ( x ) x d x = 0 sin ( x ) x d x = π 2
Hence,
F ( s ) = π 2 arctan ( s ) s = 1 s arctan ( 1 s )
Paloma Sanford

Paloma Sanford

Beginner2022-11-01Added 3 answers

Integrate by parts:
0 e s x ( 0 x sin t t d t ) d x = [ e s x s ( 0 x sin t t d t ) ] 0 + 1 s 0 e s x sin x x d x
The first term is obviously zero since the integral converges. Now consider
A ( s ) = 0 e s x sin x x d x .
We clearly have
A ( s ) = 0 e s x sin x d x .
It is well-known how to do this (parts, complex exponentials...), and we find
A ( s ) = 1 1 + s 2
Then
A ( s ) = A ( ) + s d t 1 + t 2 = arctan ( 1 s ) .
Hence we have the Laplace Transform as
1 s arctan ( 1 s ) .

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