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Help with integral

  1. Jul 6, 2007 #1
    can you solve:

    [tex]\int_0^\infty\;\frac{\ln x\cdot\left( \tan^{-1} x\right)^3}{1+x^2}\;\mathrm dx[/tex]
     
  2. jcsd
  3. Jul 6, 2007 #2

    Gib Z

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    I'm going to conjecture that it is zero.
     
  4. Jul 6, 2007 #3

    cristo

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    It's not zero, but it's not very nice either. Maple gives the answer as a sum of polylogarithms, lograithams and riemann zeta functions.
     
  5. Jul 6, 2007 #4
    really ? :frown: i was hoping the answer would have been a linear combination of [tex]\boxed{\pi^4 \ln 2}[/tex] and [tex]\boxed{\pi^2 \;\zeta{(3)}}[/tex] only
     
  6. Jul 6, 2007 #5
    I managed to get [itex]\frac{3}{64}[7\pi^2\zeta(3) - 31\zeta(5)][/itex] from Mathematica, though you might want to double check that.
     
  7. Jul 6, 2007 #6

    hey, thanks. :smile:

    [ i don't have mathematica or maple, so i am doing it by hand... i am poor... :frown: ]


    anyhow, i will also try and see if i can do

    [tex]\int_0^\infty\;\;\frac{\ln {(1+x^2)}\cdot \left( \tan^{-1} x\right)^3}{1+x^2}\;\;\mathrm dx[/tex]
     
  8. Jul 6, 2007 #7

    CompuChip

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    Did you say
    [tex]\frac{3}{64} \left(7 \pi ^2 \zeta (3)-31 \zeta (5)\right)[/tex]

    That's what I got too... question is now, why is this logical? :)
     
  9. Jul 6, 2007 #8

    CompuChip

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    Probably that's easier. Incidentally, the integral of arctan x is 1/x2 which suggests partial integration; also there are a lot of 1 + x2 in there... with a little effort you might be able to do this analytically...

    [edit] on second thought, the indefinite integral also contains a lot of PolyLogs and other creepy functions; the definite integral evaluates to
    [tex]\frac{1}{64} \left(\pi^4 \log (4)+24 \pi^2 \zeta (3)-93 \zeta(5)\right)[/tex]
    so I doubt my remarks above[/edit]
     
    Last edited: Jul 6, 2007
  10. Jul 6, 2007 #9
    Last edited by a moderator: Apr 22, 2017
  11. Jul 6, 2007 #10
    did you actually mean to say since [tex]\frac{\mathrm d}{\mathrm dx}\left( \tan^{-1} x\right) \;=\;\frac{1}{1+x^2}[/tex],
    then i could do [tex]\textbf u\;=\;\ln {(1+x^2)}\;\;\;\;\;\textbf{and}\;\;\;\;\;\textbf v\;=\;\frac{\left( \tan^{-1} x\right)^3}{1+x^2}\;\;\;?[/tex]


    but then you will get [tex]\lim_{R\to\infty}\;\frac{\pi^4}{8}\ln R[/tex] in the first part and you need to account for it somewhere along the rest of the integration process....


    PS: [tex]\int \;\tan^{-1} x\;\;\mathrm dx\;=\;x\;\tan ^{-1} x -\frac{1}{2}\;\ln {(1+x^2)}[/tex]



    oh darn ! there is that [tex]\zeta{(5)}[/tex] in the answer again ? :frown:

    thanks anyway ! :biggrin:
     
  12. Jul 6, 2007 #11

    CompuChip

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    Errr yeah. My holiday officially started today, appearantly my brain immediately shut down :grumpy:
     
  13. Jul 6, 2007 #12
    If you have to do it with algebra, have you thought about using substitution with [itex]u = \tan^{-1}(x) \implies du = \frac{1}{1+x^2}\,dx[/itex] to get [itex]\int \ln(1+x^2) \cdot \left [ \tan^{-1}(x) \right]^2\cdot u\,du[/itex]?
     
  14. Jul 6, 2007 #13
    For: [tex]\int_0^{\infty} \frac{\ln x \cdot \left( \tan^{-1} x \right)^3}{1+x^2} \ dx [/tex]
    The only idea I have so far to it use the substitution [tex]t=\tan^{-1} x[/tex] to get:
    [tex]\int_0^{\pi/2} t^3 [ \ln (\sin t) - \ln (\cos t) ] \ dt[/tex]
     
  15. Jul 6, 2007 #14

    Gib Z

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    Ahh kummer how did you change that infinite bound to pi/2, is [itex]\lim_{x\to \infty} \tan^{-1} x = \frac{\pi}{2}[/itex]? Because I was working on a solution to that last night and I got stuck at changing the bounds.

    And..
    Did you mean [tex]\int \log (1+x^2) u^3 \frac{du}{dx} dx[/tex]? Either way that doesn't work, you still have the log as a function of x, whilst integrating with respect to u.
     
    Last edited: Jul 6, 2007
  16. Jul 6, 2007 #15
    i was quite surprised to see that [tex]\zeta{(5)}[/tex] crept into the answer. :redface: i am still doing the manual calculation, things are getting quite messy, so i am taking a little break now..... but i am still very interested to see how the 5th riemann zeta comes into the picture in both those integrals...
     
  17. Jul 7, 2007 #16
    Well the general solution we discovered involved polylogarithms, and apparently [itex]\mathrm{Li}_s(1) = \zeta(s)[/itex] where [itex]\Re(s) > 1[/itex], but that still doesn't really answer the question!

    I can't even tell what I was thinking. Sorry.
     
  18. Jul 7, 2007 #17

    VietDao29

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    Well, do you know the limit:
    [tex]\lim_{x \rightarrow + \infty} \arctan (x) = \frac{\pi}{2}[/tex]?

    Since arctan(x) only returns values on the interval [tex]\left] -\frac{\pi}{2} ; \ \frac{\pi}{2}\right[[/tex], and on that interval, it's a 1-to-1 function, we also know that: [tex]\lim_{x \rightarrow \frac{\pi}{2}} \tan (x) = + \infty[/tex], so, we have: [tex]\lim_{x \rightarrow + \infty} \arctan (x) = \frac{\pi}{2}[/tex].

    You can prove that: [tex]\lim_{x \rightarrow - \infty} \arctan (x) = -\frac{\pi}{2}[/tex] in the same manner. :)
     
  19. Jul 8, 2007 #18

    Gib Z

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    Could you tell me how to derive that first limit in your post? I read something about that not being the actual limit, and that the limit does in fact not exist as the values oscillate or something like that, and the limits usually given to us, like the one in question, is the Principal Value of the limit. Although in this case I'm sure that is precisely what we want, The Principal Value of the Integral, as the integral clearly diverges otherwise.
     
  20. Jul 8, 2007 #19
    do you agree that

    [tex]\lim_{x\to+\infty}\;\tan^{-1} x\;=\;\lim_{y\to 0^{+}}\;\tan^{-1} \frac{1}{y}\;=\;\lim_{y\to 0^{+}}\;\left[ \frac{\pi}{2}\;-\;\tan^{-1} y\right][/tex]

    [tex]=\;\frac{\pi}{2}\;-\;\lim_{y\to 0^{+}}\;\sum_{j=0}^\infty\;\frac{(-1)^j\;y^{2j+1}}{2j+1}\;=\;\boxed{\frac{\pi}{2}}\;\;\;\quad (\;\;the\;\;series\;\;is\;\;valid\;\;for\;\;|y|<1\;\;)[/tex]

    [ notice how i stated it, i said [tex]\lim_{y\to 0^{+}}[/tex], not [tex]\lim_{y\to 0}\;\;[/tex] ]

    had i said [tex]\lim_{y\to 0}[/tex], then yes, the limit does not exist because if i approach 0 from the positive side, then i will get one limit, i.e. [tex]+\frac{\pi}{2}[/tex], which will be different than if i approach 0 from the negative side, where i will get a different limit, i.e. [tex]-\frac{\pi}{2}[/tex]

    so, in conclusion,

    [tex]\lim_{x\to+\infty}\;\tan^{-1} x\;=\;\lim_{y\to 0^{+}}\;\tan^{-1} \frac{1}{y}\;=\;\frac{\pi}{2}[/tex]

    and

    [tex]\lim_{x\to-\infty}\;\tan^{-1} x\;=\;\lim_{y\to 0^{-}}\;\tan^{-1} \frac{1}{y}\;=\;-\frac{\pi}{2}[/tex]
     
  21. Jul 8, 2007 #20
    I think the point Gib Z is making (correct me if I'm wrong) is that, since [itex]\tan^{-1}(x)[/itex] is multivalued, we're using the branch cut of [itex](-\frac{\pi}{2}, \frac{\pi}{2})[/itex]. This is our choice of principal values, and then the limit is indeed [itex]\lim_{x \rightarrow \infty} \tan^{-1}(x) = \frac{\pi}{2}[/itex].
     
  22. Jul 8, 2007 #21

    VietDao29

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    Well, I did some kind of proving that in the previous post. =.="

    You should note that, tan(x) itself does not have an inverse function, since it is a many-to-one function. But, like sin(x), if we restrict the interval just to be [tex]\left] -\frac{\pi}{2} ; \ \frac{\pi}{2}\right[[/tex], then, on that interval, tan(x) meets all the requirements to have an inverse. Its inverse function is defined to be arctan(x), or some may even written that as tan-1(x) (i.e, the inversion of tan(x)).

    The domain of arctan(x) is all the reals, and its image is on the interval: [tex]\left] -\frac{\pi}{2} ; \ \frac{\pi}{2}\right[[/tex]. We notice that:
    [tex]\lim_{x \rightarrow \frac{\pi}{2} ^ -} \tan (x) = + \infty[/tex], so, that mean: [tex]\lim_{x \rightarrow + \infty} \arctan (x) = \frac{\pi}{2}[/tex] (since arctan is an inverse of tan).

    Can you get this? :)
     
  23. Jul 8, 2007 #22
    It doesn't have an inverse in the typical sense of a function, however, it can be thought of as a multivalued function, and these have been studied extensively. Consider the example of [itex]x\mapstox^2[/itex]. Then its inverse is [itex]\pm \sqrt{x}[/itex]. If we choose by convention to take the positive roots, we have the inverse [itex]\sqrt{x}[/itex], and this is known as the principal square root. The same situation occurs with the inverse of [itex]\tan(x)[/itex]. The restriction of the range as we did earlier is called a branch cut, and the principal values are those which are chosen by convention as the result for a given argument.
     
  24. Jul 8, 2007 #23

    Gib Z

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    Well that was not my actual point, but you are not at fault because I was the one mistaken :) I confused it with the cases of tan (rather than arctan), sin and cos where [tex]\lim_{x\to\infty} \mbox{One of the Trig Functions}[/tex] has no actual value because of the functions periodic nature. So instead sometimes they are given principal values. But I see It was my confusion, thanks :)

    O and VietDao29, sorry about the request for a proof when in was in my face, i feel so stupid!! And thanks nasim for a longer version :P
     
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