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Calc III infinite integration

  1. Nov 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Evaluate the limit
    1 1 1
    lim ∫ ∫ ... ∫ cos^2((pi/2n)(x1 + x2 +... xn))dx1 dx2 ... dxn
    0 0 0

    2. Relevant equations
    Well, I know that we can change this using a double angle rule, so that the integrals become 1/2 + 1/2 cos (2*pi/2n)(x1 + x2 +... xn))dx1 dx2 ... dxn

    and the integral over the 1/2 just becomes 1/2, but the other side baffles me.

    3. The attempt at a solution
    My professor tried to do this, but I don't agree with his methodology. When he integrated it, he got pi/n out front, and if you keep integrating, this would go to pi^n / n^n . However, the integral should produce n^n / pi^n if I'm not mistaken, meaning this would diverge to infinity, and not go to zero like he said. Any ideas?
  2. jcsd
  3. Nov 17, 2014 #2

    Simon Bridge

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    Show the working for the definite integral over x1 first.
  4. Nov 17, 2014 #3
    Okay, so the first integral gives n/pi (sin(pi)), and the sin(pi) = 0, so you'd be integrating 0 from there on out. Is that correct?
  5. Nov 18, 2014 #4


    Staff: Mentor

    Why? ##cos^2(\pi/(2n))## is a constant as far as any of the integrations are concerned. You can bring it out of all of the integrals.

    Edit: Never mind. I didn't count enough parentheses.
    Last edited: Nov 18, 2014
  6. Nov 18, 2014 #5
    Whoops, that's my bad. All of the x terms are inside the cosine as well, and all of them are multiplied by the pi/2n, so

    cos^2( (pi/2n)*(x1 + x2 +... xn) )
  7. Nov 18, 2014 #6

    Simon Bridge

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    ...putting ##X_m=\sum_{i=m}^n x_i## (saves typing)
    $$\int_0^1 \cos \frac{\pi}{n}(x_1+X_2)\;dx_1 = \frac{n}{\pi}\big[\sin \frac{\pi}{n}(1+X_2)-\sin \frac{\pi}{n}X_2\big]$$ ... but I'm a little distracted.

    The original problem is:
    $$\lim_{n\to\infty}\int_0^1\cdots\int_0^1 \cos^2\left(\frac{\pi}{2n}X_1\right)\;dx_1\cdots dx_n$$
    Last edited: Nov 18, 2014
  8. Nov 18, 2014 #7


    Staff: Mentor

    No, my bad. What you wrote was clear, but I misread it.

  9. Nov 18, 2014 #8

    Ray Vickson

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    It might help to use the fact that ##\cos(w)## is the real part of ##e^{iw}##, so you need to integrate
    [tex] \int_0^1 \int_0^1 \cdots \int_0^1 e^{i c (x_1 + x_2 + \cdots + x_n)} \, dx_1 \, dx_2 \cdots \, dx_n,[/tex]
    where ##c = \pi/n##. The exponential factors into separate factors for each variable, and so the n-fold integration is just the product of single-variable integrals. You can pull the "real part" operator outside all the integrals (why?).
    Last edited: Nov 18, 2014
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