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How to calculate this integral

  1. May 21, 2014 #1
    1. The problem statement, all variables and given/known data

    As part of a larger problem involving integrating over a circular wedge contour in the complex plane starting at the origin, I have parametrised the integral, as is asked to do so in the question.

    2. Relevant equations

    I end up with one integral that has proved, for me, to be extremely difficult to find [tex]\lim_{R\rightarrow +\infty} {\int_0^R 1/(x^3+1)^3\,dx}[/tex]

    3. The attempt at a solution

    I have tried a variety of methods to compute this. I know, from working which uses the residue theorem, that this integral should equal [tex]{\frac{10\pi}{27\sqrt{3}}}[/tex], and have been able to compute all other parts of the question.
    I tried partial fractions, but because the denominator is of degree 9, this would take an incredibly long time, longer than I believe is necessary, because other people have completed this in fewer steps.
    I have tried integration by parts, but this does not seem to work, as I do not get the answer that I know to be correct from using the residue theorem.

    Basically, what I would appreciate is a suggestion of another method, and perhaps a overview of the initial steps, to do an integral like this.

    Any help is greatly appreciated.

    Many thanks,

    Last edited by a moderator: May 21, 2014
  2. jcsd
  3. May 21, 2014 #2
    Hi Mark! Welcome to PF!

    Can you please post the integral you are working with? It is difficult to see what you are trying to find.
  4. May 21, 2014 #3

    George Jones

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    Have you tried partial fractions? I haven't pursued this, but the denominator does factor into a product of two lower degree polynomials.
  5. May 21, 2014 #4

    Ray Vickson

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    (1) How can an integral from 0 to R be a number like ##10 \pi /(27\sqrt{3})## that does not depend on R?
    (2) Why do you say that ##x^3+1## is of degree 9? It looks like degree 3 to me.

    If you mean that the denominator is ##x^9+1##, you get an integral that is "doable", but nasty and lengthy. Maple's answer for ##\int_0^R dx/(x^9+1)## takes 10 pages to write out.
    Last edited: May 21, 2014
  6. May 21, 2014 #5

    I'm not sure how to get it to look proper, but I end up with one integral that has proved, for me, to be extremely difficult to find ∫dx/(x^3+1), where the terminals go from 0 to R, and we are integrating along the real axis.
  7. May 21, 2014 #6
    I don't think I understand "integrating along the real axis", this is probably something related to countour integration.

    What I asked was to post the exact definite integral you are dealing with, if possible, I can help you solve the problem using real methods as I don't know anything about complex analysis or countour integration.
    Last edited: May 21, 2014
  8. May 21, 2014 #7
    We need to calculate this:
    [tex]\lim_{R\rightarrow +\infinity} {\int_0^R 1/(x^3+1)^3\,dx}[\tex]
    (sorry i am still learning how to type these properly). I have used complex analysis and contour integration to get to this stage, but I think this can be treated a a real integral.
    I know the answer is meant to look similar to
    as i have already calculated what it should be via the Residue theorem, but to calculate it by parametrisation, i somehow need to evaluate this integral, and then take the limit as R goes to infinity.

    I have tried partial fractions, but i know that, as it is degree 9, the arithmetic will take ages, and I am certain it can be done faster, as I know other people who have completed this exercise.
  9. May 21, 2014 #8


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    Fixed your tex. Use [/tex] not [\tex] to close the tex statements.
  10. May 21, 2014 #9
    That's certainly better. I have come up with a method but I am sure that faster ones exist.

    Consider an easier problem first. Can you evaluate the following:
    $$\int_0^{\infty} \frac{dx}{a^3+x^3}$$
  11. May 21, 2014 #10
    Well, i would factorise it into three linear factors and then use partial fractions, integrating each fraction separately.
  12. May 21, 2014 #11
    Yep, so what do you get? You should get a nice expression. :)

    EDIT: Erm....wait, three? ##x^3+a^3=(x+a)(x^2-ax+a^2)##.
  13. May 21, 2014 #12
    Well, if we assume a > 0, then to factorise the quadratic factor we would need to introduce complex numbers. But this should be okay, because the integrand can be complex, even if the variable can only be real.

    But if I were to keep the integrand real to go along with what you're saying (which I really appreciate), I would end up with A/(x+a) + (Bx+C)/(x^2-ax+a^2), and I would evaluate A, B and C by equating coefficients.
  14. May 21, 2014 #13
    Yes, now what result do you get? Its quite straightforward once you evaluate the definite integral I ask you.
  15. May 21, 2014 #14
    Well integrating the first fraction would give me Alog|R|-Alog(0)
    But I'm having a bit of trouble with the second. It doesn't seem to require a trig substitution, and I don't seem to be able to get it to work by substitution. For example, if I take u=x^2-ax+a^2, then du=(2x-a)dx, but I can't then substitute this in easily because I will still have the B and the C, and I can't add coefficients to satisfy B=2 without making c not equal to -a and vice versa.
  16. May 21, 2014 #15
    I am unable to follow you here.
    $$\Rightarrow \frac{1}{x^3+a^3}=\frac{x^2(A+B)+x(aB-aA+C)+a^2A+aC}{x^3+a^3}$$
    So you have the following three equations:
    What do you get for A,B and C?
  17. May 21, 2014 #16
    Did you mean that [tex]a^2A=aC=1[/tex]? I assumed that you did, and If I solved these I get: [tex]A=1/3a^2 and B=-1/3a^2 and C=2/(3a)[/tex]
    Last edited: May 21, 2014
  18. May 21, 2014 #17
    Thank you very much for your response. In our course, however, we have not yet covered the gamma, beta and zeta functions, and this leads me to believe that the integral can be done using parametrisation.
  19. May 21, 2014 #18
    I meant what I wrote but you got the correct coefficients.

    So you have:
    $$\int_0^{R} \frac{dx}{a^3+x^3}=\int_0^{R} \frac{dx}{3a^2(a+x)}+\int_0^{R} \frac{2a-x}{3a^2(x^2-ax+a^2)}dx$$

    I suppose the first integral is easy to evaluate.

    To evaluate the second one, try splitting it in the following way:
    $$\int_0^R \frac{2a-x}{3a^2(x^2-ax+a^2)}\,dx=\frac{1}{6a^2}\left(\int_0^R \frac{3a}{x^2-ax+a^2}\,dx-\int_0^R \frac{2x-a}{x^2-ax+a}\,dx\right)$$

    The first one can be handled by completing the square and the second one by the substitution ##u=x^2-ax+a^2##.
  20. May 21, 2014 #19


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    On checking, I realize that I made an error in the initial substitution, so I deleted my post.

    However you do end up with a beta function, and I think that you have [tex]
    \int_0^\infty \frac{1}{(1 + x^3)^3}\,dx = \frac13 \int_0^1 u^{5/3} (1 - u)^{-2/3}\,du
    = \tfrac13 B(\tfrac 83, \tfrac 13).[/tex]
    You will need the additional result [tex]
    \Gamma(z + 1) = z \Gamma(z).[/tex]
  21. May 21, 2014 #20


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    looks pretty good. But to continue from here, he needs to know quite a few more formulas related to the beta and gamma functions. Or simply use beta(8/3,1/3) in an advanced calculator :)
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