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Homework Help: Hard integrals

  1. Dec 19, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \int_0^{\infty} \ln \left( \frac{e^x+1}{e^x-1} \right) \mbox{d}x
    [/tex]

    [tex]
    \int_0^{\infty} \frac{1}{x^n+1}\ \mbox{d}x\ \forall n >1
    [/tex]

    2. Relevant equations
    -


    3. The attempt at a solution
    I've tried IBP and separating the ln into two terms and failed. I've also tried a subtitution for exp(x)+1 and exp(x)-1 with no succes. For the second one I've tried using a trigoniometric substitution but it didn't work out.
     
  2. jcsd
  3. Dec 19, 2011 #2
    If you notice, these integrals can't be evaluated with standard integration techniques. These are called Improper Integrals. I would read up on this as I can't teach you the entire thing itself. You would need to understand the types of improper integrals as well as know the comparison theorem.
     
    Last edited: Dec 19, 2011
  4. Dec 19, 2011 #3
    Try to make the substitution:
    [tex]
    t \equiv \frac{e^x + 1}{e^x - 1}
    [/tex]

    What integral do you get?
     
  5. Dec 19, 2011 #4
    I get:

    [tex] 2 \cdot \int_1^{\infty} \frac{\ln(t)}{1+t}\ \mbox{d}t [/tex]

    but I don't see how to solve this one...
     
  6. Dec 19, 2011 #5
    Your denominator is wrong.
     
  7. Dec 19, 2011 #6

    Char. Limit

    User Avatar
    Gold Member

    Are you sure about that? I got something different.
     
  8. Dec 20, 2011 #7
    My bad:

    [tex] 2\cdot \int_1^{\infty} \frac{\ln(t)}{(t+1)(t-1)}\ \mbox{d}t [/tex]

    I can use partial fraction decomposition but I'm still stuck then. I've also tried a [tex]t =sin(\theta)[/tex] but the what?
     
    Last edited: Dec 20, 2011
  9. Dec 20, 2011 #8
    Ok, so, now your denominator is [itex]t^2 - 1[/itex]. You want it in the form [itex]e^{\alpha x} - 1[/itex]. That is why you have the additional substitution
    [tex]
    e^{\alpha x} = t^2 \Rightarrow t = e^{\frac{\alpha \, x}{2}}
    [/tex]
    where [itex]\alpha[/itex] is yet to be determined.

    What do you get now?
     
    Last edited: Dec 20, 2011
  10. Dec 20, 2011 #9
  11. Dec 20, 2011 #10
    yes, those are the necessary tricks to evaluate the integral, although I don't know why you only gave an upper bound estimate. I can't zoom in to your picture to see the details.

    The second integral can be brought to a similar form.
     
  12. Dec 21, 2011 #11
    Is the series I calclulated in the correct form? If so how can I exactly calculate:

    [tex] \sum_{n=0}^{\infty} \frac{1}{(n-\frac{1}{2} )(n-\frac{3}{2})} [/tex]
     
  13. Dec 21, 2011 #12
    [tex]
    \frac{e^{-u/2}}{1 - e^{-u}} = \sum_{n = 1}^{\infty}{e^{-\frac{u}{2}} \, (e^{-u})^{n - 1}} = \sum_{n = 1}^{\infty}{e^{-(n - \frac{1}{2}) u}}
    [/tex]

    [tex]
    \int_{0}^{\infty}{u \, e^{-(n - \frac{1}{2}) \, u} \, du} = \frac{1}{(n - 1/2)^2} \, \int_{0}^{\infty}{t \, e^{-t} \, dt}
    [/tex]

    [tex]
    \begin{array}{l}
    \sum_{n = 1}^{\infty}{\frac{1}{(n - 1/2)^2} } = \sum_{n = 1}^{\infty}{\frac{4}{(2 n - 1)^2}} \\ \\

    = 4 \, \left[ \sum_{n = 1}^{\infty}{\frac{1}{n^2}} - \sum_{n = 1}^{\infty}{\frac{1}{(2 n)^2}} \right] \\

    = 4 \, \left( 1 - \frac{1}{4} \right) \, \sum_{n = 1}^{\infty}{\frac{1}{n^2}} = 3 \, \zeta(2)
    \end{array}
    [/tex]
    where
    [tex]
    \zeta(a) \equiv \sum_{n = 1}^{\infty}{\frac{1}{n^{a}}}, \ \mathrm{Re} a > 1
    [/tex]
    is the Riemann zeta function.
     
  14. Dec 23, 2011 #13
    How do you get from the single sum with the (2n-1)^2 in the denominator to the two sums?
     
  15. Dec 23, 2011 #14

    Curious3141

    User Avatar
    Homework Helper


    Because the union of the set of all odd positive numbers and the set of even positive numbers equals the set of natural numbers. :wink:

    Sum of reciprocals of odd positives is sum of reciprocals of naturals minus sum of reciprocals of even positives.
     
    Last edited: Dec 23, 2011
  16. Dec 23, 2011 #15
    Yes offcourse thanks guys!

    For the second integral the hint is use: u =x^n, I don't understand what to do then.
     
  17. Dec 23, 2011 #16
    no, separate it from 0 to 1, and from 1 to [itex]\infty[/itex], then the fraction in the integrand:
    [tex]
    \frac{1}{x^n + 1}
    [/tex]
    has different series expansions.
     
  18. Dec 25, 2011 #17
    But if you split this integral into two geometric series how do you solve for the integral on the infinite interval?

    And how do I calculate this integral:

    [tex]
    \int_0^1 \ln(x) e^{-x} (1-x)\ \mbox{d}x
    [/tex]

    I've tried IBP bu end up with:

    [tex]
    \int_0^1 \ln(x) e^{-x} (1-x)\ \mbox{d}x = -e^{-x} (1-x) \ln(x) ] _0^1 + \int_0^1 e^{-x} \left( \frac{(1-x)}{x} +\ln(x) (-1) \right) \mbox{d}x
    [/tex]

    but what's:
    [tex]
    e^{-x} \ln(x) ] _0^1
    [/tex]
     
  19. Dec 25, 2011 #18
    No, do IBP, but make it like this:
    [tex]
    u = \ln(x), dv = e^{-x} ( 1 - x) \, dx
    [/tex]
    What is [itex]v(x)[/itex]?
     
  20. Dec 26, 2011 #19
    Ah I see it now:

    [tex] \int_0^1 \ln(x)(1-x)e^{-x}\ \mbox{d}x = \ln(x)x e^{-x} ]_0^1 - \int_0^1 e^{-x}\ \mbox{d}x = e^{-x}]_0^1 = \frac{1}{e}-1[/tex]
     
  21. Dec 26, 2011 #20
  22. Jan 3, 2012 #21
    Ok and I''m also stuck with this one:

    [tex] I = \int_{0}^{\infty} \frac{{{e^{ - x}}\left( {1 - {e^{ - 6x}}} \right)}}{{x\left( {1 + {e^{ - 2x}} + {e^{ - 4x}} + {e^{ - 6x}} + {e^{ - 8x}}} \right)}}dx [/tex]

    I suspect an smart subtitution will do(since IBP fails here, right?).

    And I'm this one is also difficult:

    [tex]I = \int\limits_0^1 {\frac{{\sin \left( {p\ln x} \right) \cdot \cos \left( {q\ln x} \right)}}{{\ln x}}}dx[/tex]

    Is tried rewriting as one sine and [tex]u= \frac{1}{2q} \ln(x) [/tex] for this one but didn't seem get anywhere but is it the correct way to do it?
     
    Last edited: Jan 3, 2012
  23. Jan 3, 2012 #22

    Char. Limit

    User Avatar
    Gold Member

    For that first one, the very first thing I'd try is a u=e^(-x) substitution.
     
  24. Jan 3, 2012 #23

    I think the sign is wrong. oops never mind. ok, to compensate, how to partial decompose ((u^3+1)(u^3-1))/(ln(u)(1+u^2+u^4+u^6+u^8)) might be a key.
     
    Last edited: Jan 3, 2012
  25. Jan 5, 2012 #24
    yes I was of course immediately thinking about u=e^{-x} the problem is then that the partial fraction decomposition gets tough or not?
     
  26. Jan 5, 2012 #25
    Hint:
    [tex]
    1 + u^2 + u^4 + u^6 + u^8 = 0
    [/tex]
    is a symmetric polynomial equation of order 4 w.r.t. [itex]v = u^2[/itex], which may be further reduced. Divide by [itex]v^2[/itex]:
    [tex]
    \left( v^2 + \frac{1}{v^2} \right) + \left( v + \frac{1}{v} \right) + 1 = 0
    [/tex]
    and, then, make a substitution [itex]w = v + 1/v[/itex]. It's square is:
    [tex]
    w^2 = v^2 + 2 + \frac{1}{v^2}
    [/tex]
    and we finally arrive at the quadratic equation:
    [tex]
    w^2 + w - 1 = 0
    [/tex]
    which has two real solutions:
    [tex]
    w_{1/2} = \frac{-1 \pm \sqrt{5}}{2}
    [/tex]
    Then, for each solution for w, we have a quadratic equation w.r.t. v, which, in turn, gives 2 solutions. You have to further take the square root to get the 8 solutions for u.

    Just to get you started, the equations for v are:
    [tex]
    2 v^2 + (1 \mp \sqrt{5}) v + 2 = 0
    [/tex]
    which has a discriminant:
    [tex]
    D = -10 \mp 2 \sqrt{5} < 0
    [/tex]
    so the solutions for v come in complex conjugate pairs.
     
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