1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laplace Transform of sin(t)/t

  1. May 11, 2010 #1

    Char. Limit

    User Avatar
    Gold Member

    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?
     
  2. jcsd
  3. May 11, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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]
     
  4. May 11, 2010 #3

    Cyosis

    User Avatar
    Homework Helper

    You're missing a minus sign and the constant of integration. To get the correct result you will have to find that constant.
     
  5. May 11, 2010 #4

    Char. Limit

    User Avatar
    Gold Member

    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.
     
  6. May 11, 2010 #5

    Cyosis

    User Avatar
    Homework Helper

    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: May 11, 2010
  7. May 11, 2010 #6

    Char. Limit

    User Avatar
    Gold Member

    And then lastly, [tex]F(s)=\frac{\pi}{2} - tan^{-1}(s)=cot^{-1}(s)[/tex], right?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laplace Transform of sin(t)/t
Loading...