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!

Homework Help: Vanishing Integral

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

    1. The problem statement, all variables and given/known data
    http://mikef.org/files/phys_4242_hw5.pdf [Broken]
    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
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 29, 2007 #2


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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 by a moderator: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook