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Explain this method for integrals (complex analysis)

  1. Dec 26, 2014 #1
    I saw this method of calculating:

    $$I = \int_{0}^{1} \log^2(1-x)\log^2(x) dx$$


    Can you take a look at M.N.C.E.'s method?

    I dont understand a few things.

    Somehow he makes the relation:

    $$\frac{4H_n}{(n+1)(n+2)^3} = \frac{\left( \gamma + \psi(-z) \right)^2}{(z+1)(z+2)^3}$$

    How is this established?

    And this I dont understand, why did he integrate it,?

    And then after he states: "At the positive integers," what is he doing with the residues. I know the residue theorem etc, but I dont understand what he is exactly doing?
  2. jcsd
  3. Dec 26, 2014 #2


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    That relation is not made. The logic:

    Consider this function f(x) someone made up: if we integrate it over a square (as described in the comments) the integral is zero. Another way to express the same integral is via its residues: their sum has to be zero.
    He then calculates the residues at -2, -1, 0 and the sum of all remaining residues (all at positive integers). The sum of all components is zero. As the sum contains ##\sum \frac {2 H_n}{(n+1)(n+2)^3}## this allows to solve the equation for this expression. You can then use that equation for the original problem.
  4. Dec 27, 2014 #3
    @mfb, thankyou very much, this was excellent help, physicsforums is great. I would like to ask a few things if you dont mind.

    Question 1) He does not define the square though? Where are the vertices?
    -I will assume that (-2, -1, 0) are INSIDE the rectangle, not on or outside.

    Question 2) How are there infinite singularities at positive integers? http://m.wolframalpha.com/input/?i=digamma(-z)&x=0&y=0
    I think I know. \digamma(-z) yields complex infinity for all positive integer values of z.
    So there are non removable singularities.

    Question 3) How does he get this:

    $$\sum_{n=1}^{\infty} Res(f, n) = \sum_{n=1}^{\infty} Res_{z=n} \frac{1}{(z+1)(z+2)^3(z-n)^2} + \frac{2H_n}{(z+1)(z+2)^3(z-n)}$$

    More specifically, why does he have (z-n) in the denominator? And where did the $H_n$ appear from?

    Thats all for now, hopefully you can guide me, thanks.
  5. Dec 27, 2014 #4


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    It is described in the smaller comments below the main post.

    I don't know how he got those residuals. Must come from the digamma function.
  6. Dec 28, 2014 #5

    Okay. mm.. I dont understand.

    When he integrated the digamma function, how does the result in a function with H_n?? Thanks
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