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Homework Help: Difficult Integral

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data

    In my calc II book, there is an example that shows how to evaluate [itex]\int\frac{1}{1+x^7}dx[/itex] as a power series and then use the result to approximate [itex]\int^{\frac{1}{2}}_{0}\frac{1}{1+x^7}dx[/itex]. I understand how to do this. The book says in the margins that calculating [itex]\int\frac{1}{1+x^7}dx[/itex] by hand is extremely difficult, but it doesn't say it's impossible. Furthermore, wolframalpha gives the integral, but doesn't provide step-by-step results. Which brings me to my question: how can you solve this integral without using power series? Either the indefinite integral or the definite integral from 0 to 1/2.

    3. The attempt at a solution

    I don't really know where to begin. There don't seem to be any helpful substitutions. If I use integration by parts, I get [itex]\frac{x}{1+x^{7}}+7\int\frac{x^{7}}{(1+x^{7})^{2}}dx[/itex] which just seems much more complicated.

    If I try to construct a right triangle with one leg equal to 1 and the other equal to [itex]x^{\frac{7}{2}}[/itex] and then use trig substitution with [itex]tanθ=x^{\frac{7}{2}}[/itex],[itex]sec^{2}θdθ=\frac{7}{2}x^{\frac{5}{2}}dx[/itex],[itex]sec^{2}θ=1+x^{7}[/itex], I can't find a way to rewrite [itex]x^{\frac{5}{2}}[/itex] ([itex]x^{\frac{5}{2}}=tan^{\frac{5}{7}}θ[/itex] ?) in terms of θ.

    I noticed that -1 is a root of the denominator so I tried to factor out x+1 hoping that I could use a partial fraction decomposition, but I just end up with [itex]\int\frac{1}{(x+1)(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1)}dx[/itex], which again, just seems to complicate it more.

    I then tried to apply what very little I know about Leibniz Integration (differentiation under the integral sign) to try to solve the definite integral, but I had no luck. I thought of using [itex]F(t)=\int^{\frac{1}{2}}_{0}\frac{e^{-(1+x^{7})t}}{1+x^{7}}dx[/itex] as my function of t, but I just end up with [itex]F'(t)=-\int^{\frac{1}{2}}_{0}e^{-(1+x^{7})t}dx[/itex], which looks like a nonelementary integral.

    Does anyone have any other ideas, either for the definite or indefinite integral? Could you point me in the right direction? Or at least just tell me that I'm in over my head.
  2. jcsd
  3. May 7, 2012 #2
    It's basically a (potentially) horrid partial fraction decomposition followed by (potentially) equally horrid trig substitutions.

    Factoring [itex]1+x^7[/itex] into a product of linear and quadratic terms is hard enough if you haven't seen the roots of unity tricks (there not that bad, but if you haven't seen it ...).

    P.S. Partial fraction decomposition will handle all rational functions. Though the computations get really nasty in most general cases. As bad as they may have seemed, the problems you were given in your calculus class were actually very nice.
    Last edited: May 7, 2012
  4. May 7, 2012 #3
    Thanks for the input. I'll need more time to look into that method and then try to use a partial fraction decomposition. If anything, I just want to see how horrifying it gets.
  5. May 7, 2012 #4
    I don't want to discourage you from looking into it and learning a bit. But just to give you a head start on the true nastiness of the problem ...

    [itex]x^7+1=(x+1)(x^2+\alpha_1x+1)(x^2+\alpha_2x+1)(x^2+\alpha_3x+1)[/itex] where [itex]\alpha_n=2\cos\frac{2\pi n}{7}[/itex].
  6. May 7, 2012 #5
    So here's what I've found so far about unity of roots:

    To find the n roots of [itex]x^{n}=1[/itex], use the formula x=[itex]1(cos(\frac{2πk}{n})+isin(\frac{2πk}{n}))[/itex], where k=0 to n-1.

    But how can I use this to factor the denominator to what you have? Does this only work for [itex]x^{n}=-1[/itex] when n is odd, as in our case? It seems to work for [itex]x^{7}=-1[/itex], x=[itex]-1(cos(\frac{2πk}{n})+isin(\frac{2πk}{n}))[/itex],k=0,1,2,3,4,5,6. Could you provide me with a good link to an article possibly because I'm having trouble finding one.
    Last edited: May 7, 2012
  7. May 7, 2012 #6
    I don't have a good reference. For me it was more about knowing how the [itex]x^n=1[/itex] case works and why it works the way that it does. Then I can generalize to other cases as needed.

    It really all boils down to Euler's formula, [itex]e^{i\theta}=\cos\theta+i\sin\theta[/itex], and the symmetries of complex roots of polynomials with real coefficients.

    Whenever [itex]z=a+ib[/itex] is a complex root of a real polynomial, its complex conjugate [itex]\bar{z}=a-ib[/itex] is also a root of that polynomial. Then [itex]x-z[/itex] and [itex]x-\bar{z}[/itex] are complex factors of the polnomial. As an exercise, foil out [itex](x-z)(x-\bar{z})[/itex] and show that this is a quadratic with real coefficients.

    This is basically how I got the quadratics that I gave previously.

    Edit: Note that for [itex]z=e^{i\theta}[/itex], [itex]\bar{z}=e^{-i\theta}[/itex]. (Think about even/odd properties of cosine/sine)
  8. May 8, 2012 #7

    So you're saying that one of the roots of [itex]1+x^{7}[/itex] is x=-1 and the other six make up three pairs of complex conjugates? Then [itex]1+x^{7}[/itex] can be factored into:



    But how then do you know the values of [itex]a_{n}[/itex] and [itex]b_{n}[/itex]? I assume this is where the roots of unity tricks come into play. I tried using your factorization [itex]x^7+1=(x+1)(x^2+\alpha_1x+1)(x^2+\alpha_2x+1)(x^2+ \alpha_3x+1)[/itex] where [itex]\alpha_n=2\cos\frac{2\pi n}{7}[/itex] but the right side doesn't seem to equal the left.

    EDIT: I think [itex]a^{2}_{n}+b^{2}_{n}[/itex] ends up being 1 since [itex]a_{n}=cos(\frac{2πn}{7})[/itex] and [itex]b_{n}=sin(\frac{2πn}{7})[/itex] and [itex]sin^{2}x+cos^{2}x=1[/itex], right? So then I can rewrite what I wrote as:


    But are you sure that the formula for [itex]a_{n}[/itex] is correct?
    Last edited: May 8, 2012
  9. May 8, 2012 #8

    If you are interested in integral in (-inf, +inf) integral of complex function gives it. z^7+1=0 has seven solution z=e^iπ/7, e^iπ3/7, z=e^iπ5/7,-1,e^iπ9/7,e^iπ11/7,e^iπ13/7.
    So you should think of three residues from three of these poles in upper or lower half of the complex plane and half of pole -1 for your integral.

    Last edited: May 8, 2012
  10. May 8, 2012 #9
    I'm going to need to learn complex analysis before I can do this.
  11. May 8, 2012 #10
  12. May 8, 2012 #11
    Hi. Using complex function
    1/(1+z^7) = Σk Ak / (z - e^iπ(2k+1)/7)
    Integrating this,
    ∫1/(1+z^7)dz = Σk Ak ln (z - e^iπ(2k+1)/7)
    same as wolfram, isn't it. Regards.
    Last edited: May 8, 2012
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