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The limit of some integral

  1. Jul 31, 2009 #1
    Dear all,
    I want to calculate the following integral

    [tex]
    \int_{-\infty}^0 dk \frac{k\left(\frac{k^2-m^2}{k}\cos\frac{2(x M - k)c_0}{m y} + m\sin\frac{2(x M - k)c_0}{m y} + \frac{k^2+m^2}{2k}\right)}{\sinh^2\frac{(x M - k)\pi}{2my}((k^2 - m^2)^2 + 4 k^2 m^2 y^2)}
    [/tex]

    in the limit [tex]y\to 1[/tex] to examine the small x regime (x > 0, x << 1). However, [tex]c_0[/tex] is given by

    [tex]
    c_0 = \frac{1}{2}\operatorname{arctanh}y
    [/tex]

    so it diverges in the limit [tex]y\to 1[/tex]. But then I would state that we may neglect the cosine and sine terms since they oscillate so rapidly that there contribution to the integral vanishes. My professor however, with whom I discussed this matter, says I am not eligible to do that since I want to examine the low-x regime where I get a pole in the limit [tex]x\to 0[/tex]. Than my arguing would not be true.

    I told him I would try to give this one a rigourious mathematical treatment. But then 1) I can't see what is wrong with my arguments since I am not examing x = 0 but only small but non-zero x where there is no pole and 2) it looks so obvious to me that I don't really know how to treat this on solid mathematical grounds.

    Thats why I would really appreciate a discussion about this integral in the limit [tex]y\to 1[/tex]. Hopefully some of you has some idea how to treat this.

    A big thanks in advance!
    Blue2script
     
  2. jcsd
  3. Aug 4, 2009 #2
    Yowzers, quite an integral.

    If you're assuming small x...

    Mx-k == -k


    As long as Mx <<< k
     
  4. Aug 5, 2009 #3

    Mute

    User Avatar
    Homework Helper

    I don't know if it would help any, but you could combine the cosine and sine using the identity

    [tex]A\cos\phi + B\sin\phi = \sqrt{A^2 + B^2}\sin\left(\phi + \tan^{-1}\left(\frac{A}{B}\right)\right)[/tex]

    (note: if A/B < 0, then a phase factor of pi must be added)
     
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