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Proof: Compare two integral(Please look at my surgested proof)

  1. Feb 27, 2008 #1
    This is a repost, the reason I feared that people who miss the the original.

    1. The problem statement, all variables and given/known data

    Looking at the Integral

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

    prove that [tex]a_n \geq a_{n+1}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Here is my now proof:

    the difference between the two integrals, we seek to show:

    [tex]$\forall n\in\mathbb{N}:\int_0^\pi\left({\sin t\over t+n\pi}-{\sin t\over t+(n+1)\pi}\right)\,dt\ge 0[/tex]

    Common denominator:

    [tex]$=\int_0^\pi\left({\sin t((t+(n+1)\pi)-(t+n\pi))\over (t+n\pi)(t+(n+1)\pi)}\right)\,dt[/tex]

    [tex]$=\pi\int_0^\pi\left({\sin t\over (t^2+(2n+1)\pi t+(n^2+n)\pi)}\right)\,dt[/tex]

    From here we can use the fact that the denominator on the half of the interval where sinus is negative is larger than the denominator of each and every other corresponding point on the other half of the interval, so the whole integral must be positive.

    q.e.d.

    How does it look now?

    Sincerely Maria.
     
  2. jcsd
  3. Feb 27, 2008 #2

    Dick

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    But sin(t) isn't negative anywhere on [0,pi]. That makes life a lot easier.
     
  4. Feb 27, 2008 #3
    Looks good.
     
  5. Feb 27, 2008 #4
    Hi Dick,

    So I change my conclusion.

    Since sin(t) is positive on the interval [0,pi] then the who integral most be positive and thusly convergent.

    Does this sound better?

    Sincerely

    Maria
     
  6. Feb 27, 2008 #5

    Dick

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    If everything is nonnegative all you need is that t+n*pi<t+(n+1)*pi. So 1/(t+n*pi)>1/(t+(n+1)*pi). It becomes pretty obvious that your integral is positive.
     
  7. Feb 27, 2008 #6
    And thusly it converges?

    Sincerely
    Maria.
     
  8. Feb 27, 2008 #7

    Dick

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    It converges because the integrand and domain of integration are bounded, right?
     
  9. Feb 27, 2008 #8
    Here is my original proof:

    Looking at the Integral

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

    prove that [tex]a_n \geq a_{n+1}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Here is my now proof:

    the difference between the two integrals, we seek to show:

    [tex]$\forall n\in\mathbb{N}:\int_0^\pi\left({\sin t\over t+n\pi}-{\sin t\over t+(n+1)\pi}\right)\,dt\ge 0[/tex]

    Common denominator:

    [tex]$=\int_0^\pi\left({\sin t((t+(n+1)\pi)-(t+n\pi))\over (t+n\pi)(t+(n+1)\pi)}\right)\,dt[/tex]

    [tex]$=\pi\int_0^\pi\left({\sin t\over (t^2+(2n+1)\pi t+(n^2+n)\pi)}\right)\,dt[/tex]

    The integral is therefore non-negativ and therefore

    [tex]t+n*pi<t+(n+1)*pi[/tex] and [tex]/(t+n*pi)>1/(t+(n+1)*pi).[/tex] thus the integral is positive.

    The integral is therefore bounded since the integrand is continous which makes the original in-equality true.

    q.e.d.

    Is it on the mark now?

    Thanks in advance

    Sincerely Yours
    Maria.
     
  10. Feb 27, 2008 #9

    Dick

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    i) Your 'therefores' are somewhat backwards. sin(t)/(t+n*pi)>sin(t)/(t+(n+1)*pi) implies the integrand is positive. The integrand being positive DOESN'T imply the inequality. So, ii) you don't NEED a common denominator anymore. All the parts are there, you could make the phrasing clearer.
     
  11. Feb 27, 2008 #10
    Okay and thank your for your answer,

    Here we go again:

    proof:

    By applying the limit

    [tex]sin(t)/(t+n*pi)>sin(t)/(t+(n+1)*pi)[/tex] the integrand becomes positive for all n. The integral is positive and inequality is thusly true.

    How about now?

    Best Regards.
    Maria.
     
  12. Feb 27, 2008 #11

    Dick

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    It's not a 'limit', it's an inequality. But good enough.
     
  13. Feb 27, 2008 #12
    Last edited: Feb 27, 2008
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