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Proving the equality of two integrals

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove

    [tex]\int_{n\pi}^{(n+1)\cdot \pi} \frac{sin(x)}{x} dx = (-1)^n \cdot a_n[/tex]

    where [tex]a_n = \int_{0}^{\pi} \frac{sin(x)}{n\cdot pi x} dx[/tex]

    and [tex]n \geq 0[/tex]


    3. The attempt at a solution

    My Question is how I do prove that the left side equals the right side? Writting their respective Riemann sums?

    Sincerely Yours
    Hummingbird
     
  2. jcsd
  3. Feb 26, 2008 #2

    morphism

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    Are you sure you wrote it out correctly? The second integrand isn't defined when n=0, and when n=1 they're certainly not equal.

    Did you perhaps mean to write
    [tex]a_n = \int_0^{\pi} \frac{\sin x}{x - n\pi} \, dx[/tex]
    instead?
     
  4. Feb 26, 2008 #3
    Hi morphism,

    I meant to write

    [tex]a_n = \int_0^{\pi} \frac{\sin x}{x + n\pi} \, dx[/tex].

    Do I write out each side as a series or sum maybe? I would much appriciate a hint if possible:)

    Because

    I [tex]\int_{n\pi}^{(n+1)\cdot \pi} \frac{sin(x)}{x} dx = sin(x)/x |_{n\pi}^{({n+1})\cdot \pi}[/tex]

    Sincerely Yours
    Hummingbird
     
    Last edited: Feb 26, 2008
  5. Feb 26, 2008 #4

    morphism

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    Sorry - that's what I meant to write as well!

    Try the substitution [itex]y=x-n\pi[/itex].
     
  6. Feb 26, 2008 #5
    Just be clear you mean I need to find the n anti-derivate of the integral and prove that its the same as the integral on right?

    If I do what you ask then [itex]y=x+n\pi[/itex] then I get that

    [tex]\frac{1}{y} \cdot \int_{0}^{\pi} sin(x) dx[/tex] where [tex]\frac{dy}{dx} = 1[/tex]
     
    Last edited: Feb 26, 2008
  7. Feb 26, 2008 #6

    morphism

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    I meant for the substitution to be used with the first integral you posted. Is that what you did? I don't understand how you got a \frac{1}{y} on the outside.
     
  8. Feb 26, 2008 #7
    Sorry then

    [tex]\int_{0}^{\pi} \frac{sin(x)}{y} dx[/tex]


    I get the anti-derivative of the above to be

    [tex]\frac{-cos(x)}{n \dot \pi+x}[/tex]
     
  9. Feb 26, 2008 #8

    morphism

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    It seems like you aren't applying the substitution properly.

    The integral in question is
    [tex]\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x} \, dx.[/tex]

    If we let [itex]y=x-n\pi[/itex] then it becomes
    [tex]\int_{0}^{\pi} \frac{\sin (y+n\pi)}{y+n\pi} \, dy.[/tex]

    (Notice the change of variables!)

    Using a certain identity to simplify [itex]\sin (y+n\pi)[/itex] will wrap things up.
     
  10. Feb 26, 2008 #9
    Excuse me for asking you mean the unit circle identity where the signe in front of sinus changes constantly as n increases?
     
  11. Feb 26, 2008 #10

    morphism

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    I mean sin(x+y)=sin(x)cos(y)+sin(y)cos(x).
     
  12. Feb 26, 2008 #11
    [tex]\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dx[/tex]

    Excuse me for asking but how does that simplify things?
     
    Last edited: Feb 26, 2008
  13. Feb 26, 2008 #12
    just so you know, you are integrating with respect to y now!
     
  14. Feb 26, 2008 #13
    Thank You

    [tex]\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dy[/tex]

    Do I factor the above or what do I do ?

    SR

    HM
     
  15. Feb 26, 2008 #14
    well, i am not sure, but i think that factoring would really simplify things, because noteice that sin(npi), and cos(npi) are sonstants so you would end up with sth similar to
    [tex]\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dy=sin(n\cdot\pi)\int_0^{\pi}\frac{cos(y)}{y+n\pi}dy+cos(n\pi)\int_0^{\pi}\frac{sin(y)}{y+n\pi}dy= (-1)^{n} a_n[/tex]

    now [tex]a_n = \int_0^{\pi} \frac{\sin y}{y + n\pi} \, dy[/tex]
     
    Last edited: Feb 26, 2008
  16. Feb 26, 2008 #15
    Hi and thank you for your answer,

    I can see that in order to achieve the result I need to get rid of this term here

    [tex]sin(n\cdot\pi)\int_0^{\pi}\frac{cos(y)}{y+n\pi}dy[/tex]

    Changing this into a sum would that allow me to get rid of this?

    It can't simple be that I set n = 0 zero in the first part of the integral to get rid of it?

    Sincerely
    Maria the (Hummingbird)
     
    Last edited: Feb 26, 2008
  17. Feb 26, 2008 #16
    If n is an integer, what can you say about the value of [itex]\sin n\pi[/itex]? What about the value of [itex]\cos n\pi[/itex]?
     
  18. Feb 26, 2008 #17
    Yep, and end of story!>.<
     
  19. Feb 26, 2008 #18
    If n icreases [itex]\sin n\pi[/itex] always be zero while [itex]\cos n\pi[/itex] will have its sign in fron change from - to plus everytime n increases!


    Sincerely Maria.
     
    Last edited: Feb 26, 2008
  20. Feb 26, 2008 #19
    Yes [tex]sin(n\pi)=0,while----

    cos(n\pi)=(-1)^{n}[/tex]
    You are done, have another look at post #14
     
    Last edited: Feb 26, 2008
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