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Vanishing Integral

  1. Dec 28, 2007 #1
    [SOLVED] Vanishing Integral

    1. The problem statement, all variables and given/known data
    When does the integeral at the bottom of page 2 vanish? He says it vanishes when n'_y and n'_y have oppositive parity, but I think it always vanishes because n'_y - n'_y and n'_y + n'_y always have the same parity, so the integral over

    cos(pi (n'_y - n'_y) y) and cos(pi (n'_y + n'_y) y)

    should be the same.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 29, 2007 #2


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    The anti-derivative of cos is sin, which is 0 at any multiple of [itex]\pi[/itex]. At the upper limit, the b in the denominator is cancelled by the upper limit b, leaving an integeer multiple of [itex]\pi[/itex]. Of course, at the lower limit, 0, sin(0)= 0.
  4. Dec 29, 2007 #3
    Notice there is a y/2 in front. Could you explain what your answer means for the result of the integral? It IS always 0, correct? So, the last sentence is wrong?
  5. Dec 29, 2007 #4


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    You are right, I missed the y/2 completely! Okay, do an "integration by parts". Let u= y and dv= (cos(py)- cos(q))dy. Then du= dy, v= -(sin(py)/p+ sin(qy)/q).

    The integeral is "[itex]uv\left|_0^b- \int_0^b vdu[/tex]" . As before The first term will be 0 at both y= 0 and y= b. Now the problem is just
    [tex]\int_0^v sin(py)/p+ sin(qy)/q)dy= -cos(py)/p^2- cos(qy)/q^2[/tex]
    evaluated between 0 and b. Now, if [itex]n_y[/itex] and [itex]n_y'[/itex] differ by an even number[/b] (i.e. are either both even or both odd) those reduce to cosine at 0 and an even multiple of [itex]\pi[/itex] and so the difference is 0. That is the reason for the condition "unless [itex]n_y' = n_y[/itex]+ an odd positive integer".
    Last edited: Dec 29, 2007
  6. Dec 29, 2007 #5
    I think it should be

    [tex]\int_0^v sin(py)/p- sin(qy)/q)dy= -cos(py)/p^2+ cos(qy)/q^2[/tex]

    but now I see why it is 0 unless n_y and n'_y have opposite parity. Thanks.
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