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Logarithm, identity

  1. Oct 27, 2009 #1
    I read the following expression in a book:

    [tex] \int_{-\infty}^{\infty} \dfrac{1}{t(1-t)} \log \left| \dfrac{t^{2} q^{2}}{(p-tq)^{2}} \right| ~ dt = - \pi^{2} [/tex]

    p and q are both timelike four-vectors, so p², q² > 0

    This integral was solved by using the identity

    [tex] \lim_{s \to \infty} \int_{-s}^{s} \dfrac{\log \vert t-a \vert}{t-b} ~ dt = \dfrac{\pi^{2}}{2} sign(b - a) [/tex]

    But I don't know how I can apply this identity to the integral above. I don't find any substitution or anything else to 'convert' the integral
    in such a shape where I can use these formula.

    Could anyone help me please???
     
  2. jcsd
  3. Oct 27, 2009 #2

    tiny-tim

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    Hi parton! :smile:

    Try splitting the 1/t(1-t) in partical fractions, and the log into the difference of two logs, and then apply the identity several times. :wink:
     
  4. Oct 28, 2009 #3
    Hi tiny-tim ! Thanks for your answer :smile:

    I tried to split the integral as you suggested:

    [tex] \int_{-\infty}^{\infty} \dfrac{1}{t(1-t)} \log \left| \dfrac{t^{2} q^{2}}{(p-tq)^{2}} \right| ~ dt = \int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \left[ \log \vert t^{2} q^{2} \vert - \log \vert (p-tq)^{2} \vert \right] ~ dt [/tex]

    For the first part, I've got:

    [tex] \int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert t^{2} q^{2} \vert \right] ~ dt= - \pi^{2} [/tex]

    Here, I decomposed the logarithm again, i.e.

    [tex] \log \vert t^{2} q^{2} \vert^{2} = 2 \log \vert t \vert + \log \vert q^{2} \vert [/tex]

    I just used the identity for computing
    [tex] - 2 \int_{-\infty}^{\infty} \dfrac{1}{t-1} \log \vert t \vert = - \pi^{2} [/tex]

    The other integrals coming from this part will vanish, for example:

    [tex] 2 \int_{-\infty}^{\infty} \dfrac{1}{t} \log \vert t \vert ~ dt = \left[ \log \left( \vert t \vert \right) ^{2} \right]_{-\infty}^{\infty} = 0 [/tex]

    I argued that: [tex] \left[ \log \left( \vert t \vert \right) ^{2} \right]_{-s}^{s} = 0 [/tex], so it has also to be valid for s--> infinity. But I'm not sure wheter this argument is correct.

    Nevertheless, I don't know how to compute the second part of the originial integral:

    [tex] - \int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert (p-tq)^{2} \vert ~ dt [/tex]

    There is this nasty square appearing in the logarithm, and now I don't know how to continue :confused:

    To you know some way to solve this???
     
  5. Oct 28, 2009 #4

    Hurkyl

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    Don't you know something about log xe?
     
  6. Oct 28, 2009 #5
    ok, maybe I could simply write:

    [tex] -\int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert (p-tq)^{2} \vert ~ dt = -2 \int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert p -t q \vert ~ dt [/tex]

    But how do I solve that? p and q are four-vectors, so I can't simply apply a substitution !?
     
  7. Oct 28, 2009 #6

    tiny-tim

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    Hi parton! :wink:

    The difficulty is that p and q are 4-vectors.

    But (p - tq)2 is a quadratic in t (with coefficients which are combinations of the coordinates of p and q), so it must be of the form (t - a)(t - b), where (p - aq)2 = (p - bq)2 = 0.

    Does that help? :smile:
     
  8. Oct 28, 2009 #7
    Yes, it does help, thanks a lot :smile:

    So, I wrote:

    [tex] (p-tq)^{2} = (t-t_{1}) (t-t_{2}) [/tex]

    where [tex] t_{1,2} = \dfrac{1}{q^{2}} \left[ pq \pm \sqrt{ (pq)^{2} - p^{2} q^{2} } \right] [/tex]

    And I've got:

    [tex] - \int_{-\infty}^{\infty} \dfrac{1}{t} \left[ \log \vert t - t_{1} \vert + \log \vert t - t_{2} \vert \right] = - \dfrac{\pi^{2}}{2} \left[ sign(-t_{1}) + sign(-t_{2}) \right] = 0 [/tex]

    This is of course equal to zero, because of the opposite signs of [tex] t_{1} [/tex] and [tex] t_{2} [/tex].

    But the next one isn't so easy:

    [tex] + \int_{-\infty}^{\infty} \dfrac{1}{t-1} \left[ \log \vert t - t_{1} \vert + \log \vert t - t_{2} \vert \right] = \dfrac{\pi^{2}}{2} \left[ sign(1-t_{1}) + sign(1-t_{2}) \right] [/tex]

    Ok, I know that if [tex] t_{2} < 0 \Rightarrow 1-t_{2} >0 \Rightarrow sign(1-t_{2}) = 1 [/tex]

    But what is [tex] sign(1-t_{1}) [/tex] ?
     
  9. Oct 28, 2009 #8
    sorry, I made a mistake.

    The equation above is not equal to zero.

    If I choose a frame where [tex] p = (p_{0}, \vec{0}) [/tex] than I have:

    [tex] t_{1,2} = \dfrac{p_{0}}{q^{2}} \left[ q^{0} \pm \vert \vec{q} \vert \right] < 0 [/tex] because of [tex] q^{2} = q_{0}^{2} - \vec{q} \, ^{2} > 0 [/tex] and [tex] q_{0} < 0 [/tex].

    So I obtain:

    [tex] - \int_{-\infty}^{\infty} \dfrac{1}{t} \left[ \log \vert t - t_{1} \vert + \log \vert t - t_{2} \vert \right] ~ dt = - \dfrac{\pi^{2}}{2} \left[ sign(-t_{1}) + sign(-t_{2}) \right] = - \pi^{2} [/tex]

    and

    [tex] + \int_{-\infty}^{\infty} \dfrac{1}{t-1} \left[ \log \vert t - t_{1} \vert + \log \vert t - t_{2} \vert \right] ~ dt = \dfrac{\pi^{2}}{2} \left[ sign(1-t_{1}) + sign(1-t_{2}) \right] = \pi^{2} [/tex]

    and finally:

    [tex] -\int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert (p-tq)^{2} \vert ~ dt = 0 [/tex]

    Is anything wrong in my argumentation or is it correct?
     
    Last edited: Oct 28, 2009
  10. Oct 29, 2009 #9
    By the way,

    does anyone know where I can find such identities like

    [tex] \lim_{s \to \infty} \int_{-s}^{s} \dfrac{\log \vert t-a \vert}{t-b} ~ dt = \dfrac{\pi^{2}}{2} sign(b - a) [/tex]

    Is there any book where such formulas are "collected"? I looked in different tables of integrals, but couldn't find them.
     
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