Laplace Transform of sin(t)/t

  • #1
Char. Limit
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Just out of curiosity, I've been looking into the Laplace transform and noticed that...

[tex]\mathcal{L}(sin(t))=\frac{1}{s^2+1}=\frac{d}{ds}(tan^{-1}(s))[/tex]

So, I was wondering if the Laplace transform of the sinc function was the inverse tangent function. In LaTEX...

[tex]\mathcal{L}(sinc(t))=tan^{-1}(s)[/tex]

In short, is the above a true statement?
 

Answers and Replies

  • #2
gabbagabbahey
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Close, I think you get [itex]\cot^{-1}(s)=\frac{\pi}{2}-\tan^{-1}(s)[/itex] when you do the integration properly.

[tex]\mathcal{L}\left[\frac{f(t)}{t}\right]=\int_s^\infty F(\sigma)d\sigma[/tex]
 
  • #3
Cyosis
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You're missing a minus sign and the constant of integration. To get the correct result you will have to find that constant.
 
  • #4
Char. Limit
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Wait, constant of integration? I thought that in a definite integral, the constant of integration would cancel out...

And yeah, I see that I got something wrong there gabba. I integrated from 0 to s, instead of from s to infinity. Doing it again, you're right. Here's the true statement, I guess:

[tex]\mathcal{L}(sinc(t))=cot^{-1}(s)[/tex]

I love it when I put in a trig function and get out a trig function.
 
  • #5
Cyosis
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Char.Limit said:
Wait, constant of integration? I thought that in a definite integral, the constant of integration would cancel out...

Yes I was under the impression you did something else than using the identity in post #2 namely.

I thought you had noticed that if

[tex]
f(s)=\int_0^\infty \frac{\sin (t)}{t} e^{-s t} ds
[/tex]

then

[tex]
f'(s)=-\int_0^\infty \sin( t )e^{-s t} ds = - \frac{d}{ds} \arctan s
[/tex]

from this it follows that

[tex]
f(s)=-\arctan s+c
[/tex]

but [itex]f(\infty)= 0[/itex] therefore [itex]c=\pi/2[/itex] and finally [itex]f(s)=\pi/2-\arctan s=\text{arccot} s[/itex]

Edit: Whoops this post was a bit premature. I thought you finished in post #4, but seeing as you finished while I was fixing my tex I won't edit this post.
 
Last edited:
  • #6
Char. Limit
Gold Member
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And then lastly, [tex]F(s)=\frac{\pi}{2} - tan^{-1}(s)=cot^{-1}(s)[/tex], right?
 

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