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Math Help - another challenging integral

  1. #1
    Super Member Random Variable's Avatar
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    another challenging integral

    Most of us know that  \int^{\pi/2}_{0} \ln (\sin x) \ dx = \int^{\pi/2}_{0} \ln (\cos x) \ dx = - \frac{\pi}{2} \ln 2

    But what about  \int^{\pi/2}_{0} \ln (\sin x) \ln (\cos x) \ dx ?

    The worked solution I've seen is quite confusing.
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  2. #2
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    by parts integration

    Quote Originally Posted by Random Variable View Post
    Most of us know that  \int^{\pi/2}_{0} \ln (\sin x) \ dx = \int^{\pi/2}_{0} \ln (\cos x) \ dx = - \frac{\pi}{2} \ln 2

    But what about  \int^{\pi/2}_{0} \ln (\sin x) \ln (\cos x) \ dx ?

    The worked solution I've seen is quite confusing.
    we gonna integrate the given expression by parts once taking u=ln(sinx),v=ln(cosx) and once u=ln(cosx),v=ln(sinx). for both times we will get "a undefined part" which we will eleminate from the 2 obtained equations and get a defined answer!Easy!!
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Pulock2009 View Post
    we gonna integrate the given expression by parts once taking u=ln(sinx),v=ln(cosx) and once u=ln(cosx),v=ln(sinx). for both times we will get "a undefined part" which we will eleminate from the 2 obtained equations and get a defined answer!Easy!!
    Do you mean  u = \ln(\cos x) and  dv= \ln( \sin x) dx ? But how can you even do that when  v doesn't have an elementary antiderivative?
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  4. #4
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    Opinion

    I convert this integral to another (unknown) one , not sure if it is useful .

    I consider the integral

     \int_0^{\frac{\pi}{2}} \ln(\sin(x))^2~dx

     = \int_0^{\frac{\pi}{2}}  \ln(\sin(2x))^2 ~dx

     = \int_0^{\frac{\pi}{2}}  [ \ln2 + \ln(\sin(x)) + \ln(\cos(x)) ]^2 ~dx

    Since  \int_0^{\frac{\pi}{2}}\ln(\sin(x))^2~dx = \int_0^{\frac{\pi}{2}}\ln(\cos(x))^2~dx  and

     \int_0^{\frac{\pi}{2}}\ln(\sin(x))~dx = \int_0^{\frac{\pi}{2}}\ln(\cos(x))~dx

    I find that  \int_0^{\frac{\pi}{2}}\ln(\sin(x))\ln(\cos(x))~dx = \frac{3\pi}{4}\ln^2(2) - \frac{1}{2} \int_0^{\frac{\pi}{2}} \ln(\sin(x))^2~dx

    It is still an unknown integral but I can represent it by infinite series .

    I am curious if this integral can be expressed by  \pi , \ln 2  somthing ...
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  5. #5
    Super Member Random Variable's Avatar
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    You end up finding the value of both integrals.


    Let  I =  \int^{\pi/2}_{0} \Big(\ln (\sin x)\Big)^{2} \ dx

    and  J = \int^{\pi/2}_{0} \ln (\sin x) \ln (\cos x) \ dx

    You should find that  2I+2J = I + \frac{3 \pi}{2} (\ln 2)^{2}

    and  2I-2J = \frac{\pi^{3}}{8} (I'm a bit confused by this part.)

    and then just solve simultaneously for I and J
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  6. #6
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    I got it !

    I first introduce two lemmas :

     \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^3} = \frac{\pi^3}{32}

     \int_0^1 x^{r}(\ln x)^n ~dx = \frac{ (-1)^n n! }{ (r+1)^{n+1} }

    The proofs are left to you , they should be easy . Note that the first one might be proved using Fourier Analysis, I think .


    J = \int^{\pi/2}_{0} \ln (\sin x) \ln (\cos x) \ dx

    Use  4xy = (x+y)^2 - (x-y)^2

     J = \frac{1}{4} \int^{\pi/2}_{0}[ \ln(\sin x \cos x)^2 - \ln(\tan x)^2 ]~dx

     = \frac{1}{4} \int^{\pi/2}_{0}[ ( \ln(\sin 2x) - \ln 2 )^2 - \ln(\tan x)^2 ]~dx

    For single-valued function  f(x) , we have  \int_0^{\pi/2} f(\sin 2x)~dx = \int_0^{\pi/2} f(\sin x)~dx

    So we have ,

     J = \frac{1}{4} \int^{\pi/2}_{0}[ ( \ln(\sin x) - \ln2 )^2 - \ln(\tan x)^2 ]~dx

     4J = \int^{\pi/2}_{0}( \ln(\sin x) - \ln2 )^2 ~dx - \int^{\pi/2}_{0} \ln(\tan x)^2 ~dx

    Sub  \tan x = t for the second integral ,

     4J = I + \frac{3\pi (\ln2)^2}{8} - \int_0^{\infty}\frac{(\ln x)^2}{ 1 + x^2}~dx

    Consider the integral \int_0^{\infty}\frac{(\ln x)^2}{ 1 + x^2}~dx

     = \left( \int_0^1 + \int_1^{\infty} \right) \frac{(\ln x)^2}{ 1 + x^2}~dx

    Sub. x = 1/t in the second one , we can see that the integral equals to :

    2\int_0^1 \frac{(\ln x)^2}{ 1 + x^2}~dx

     = 2 \sum_{k=0}^{\infty} \int_0^1 (-1)^k x^{2k} (\ln x)^2~dx

     = 2 \sum_{k=0}^{\infty} \frac{ (-1)^k 2}{ (2k+1)^3} = \frac{\pi^3}{8}

    Also from my previous post , we obtain ( use your symbols )

     2J + I = \frac{3\pi (\ln2)^2}{2}

     4J = I +\frac{3\pi (\ln2)^2}{2} - \frac{\pi^3}{8}

    Summation gives

     6J = 3\pi(\ln2)^2 - \frac{\pi^3}{8}

     J = \frac{\pi(\ln2)^2}{2} - \frac{\pi^3}{48}

    We also obtain a by-product  I = \frac{\pi(\ln 2)^2}{2} + \frac{\pi^3}{24}

    Just after the last line , I thought can we generalize the integral , power up to n ? I really want to go through it but i won't have much time since i am busy on my school examination .
    Last edited by simplependulum; April 19th 2010 at 05:13 AM.
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  7. #7
    Super Member Random Variable's Avatar
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    Thanks. They just stated that  2 \int_0^{1}\frac{(\ln x)^2}{ 1 + x^2}~dx = \frac{\pi^{3}}{8} without any explanation. The overall proof that they give is slightly different than yours, but not different enough to state it here.


    Anyways, you can prove
    \int_0^1 x^{r}(\ln x)^n ~dx = \frac{ (-1)^n n! }{ (r+1)^{n+1} } by doing the following:


      \int_{0}^{1} x^{r} (\ln x)^{n} \ dx = -\int_{\infty}^{0}  e^{-tr}(-t)^{n}e^{-t} \ dt (let t = -\ln(x) )

     = (-1)^{n} \int_{0}^{\infty} e^{-t(r+1)}t^{n} \ dt =(-1)^{n} \frac{1}{r+1} \frac{1}{(r+1)^{n}}\int_{0}^{\infty} e^{-u}  u^{n} \ du = ( let  u = t(r+1) )

     = \frac{(-1)^{n}}{(r+1)^{n+1}} \Gamma(n+1) = \frac{(-1)^{n}n!}{(r+1)^{n+1}}
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  8. #8
    Super Member Random Variable's Avatar
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    I can't think of the the appropriate Fourier series for the first lemma.
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  9. #9
    Super Member Random Variable's Avatar
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    I found a proof, but it's a bit convoluted.

    The Fourier series of  |x| on  (-\pi, \pi] is  |x| = \frac{\pi}{2} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos (2n+1)x}{(2n+1)^{2}}

    now integrate both sides from 0 to x

     \frac{x^{2}}{2} = \frac{\pi}{2} x - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\sin (2n+1)x}{(2n+1)^{3}}

    so  \sum_{n=0}^{\infty} \frac{\sin (2n+1)x}{(2n+1)^{3}} =  \frac{\pi x}{8}(\pi-x)

    now let  x = \frac{\pi}{2}

    then  \sum_{n=0}^{\infty} \frac{\sin (2n+1) \frac{\pi}{2} }{(2n+1)^{3}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{3}} = \frac{\pi^{3}}{32}
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  10. #10
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    Quote Originally Posted by Random Variable View Post
    I found a proof, but it's a bit convoluted.

    The Fourier series of  |x| on  (-\pi, \pi] is  |x| = \frac{\pi}{2} - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\cos (2n+1)x}{(2n+1)^{2}}

    now integrate both sides from 0 to x

     \frac{x^{2}}{2} = \frac{\pi}{2} x - \frac{4}{\pi} \sum_{n=0}^{\infty} \frac{\sin (2n+1)x}{(2n+1)^{3}}

    so  \sum_{n=0}^{\infty} \frac{\sin (2n+1)x}{(2n+1)^{3}} = \frac{\pi x}{8}(\pi-x)

    now let  x = \frac{\pi}{2}

    then  \sum_{n=0}^{\infty} \frac{\sin (2n+1) \frac{\pi}{2} }{(2n+1)^{3}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{3}} = \frac{\pi^{3}}{32}

    The proof by expanding Fourier series I think is too general and a bit boring , also hard to memorize . I know Euler gave a general but not boring method to solve this kind of series but i forgot it ( it is hard to memorize too ) . Once i find out the method , i post it here immediately , i think we may find it inspiring .
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  11. #11
    Super Member Random Variable's Avatar
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    If you're still looking for \int^{\pi/2}_{0} \Big(\ln (\sin x)\Big)^{n} \ dx , Maple gives the following results:

     I(3) = \int^{\pi/2}_{0} \Big(\ln (\sin x)\Big)^{3} \ dx = - \frac{3 \pi}{4} \zeta (3) - \frac{\pi^{3}}{8} \ln 2 - \frac{\pi}{2} (\ln 2)^{3}


     I(4) = \int^{\pi/2}_{0} \Big(\ln (\sin x)\Big)^{4} \ dx = 3 \pi \ln 2 \zeta (3) + \frac{\pi}{2} (\ln 2)^{4} + \frac{\pi^{3}}{4} (\ln 2)^{2} + \frac{19 \pi^{5}}{480}
    Last edited by Random Variable; April 20th 2010 at 12:30 AM.
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