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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
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