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    Find I = int(ln(sin(pix)dx) (x: 0-->1)
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    nice... watch this space...

    edit: decided it can't be done by parts...
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    Use the derivation of the dilogarithmic function.

    Euclid.
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    I
    = (int from 0 to 1) ln(sin(pi x)) dx
    = 2(int from 0 to 1/2) ln(sin(pi x)) dx
    = 2(int from 0 to 1/2) ln(cos(pi x)) dx

    2I
    = 2(int from 0 to 1/2) ln(sin(pi x)) + ln(cos(pi x)) dx
    = 2(int from 0 to 1/2) ln[sin(pi x)cos(pi x)] dx
    = 2(int from 0 to 1/2) ln[(1/2)sin(2pi x)] dx
    = 2(int from 0 to 1/2) ln(sin(2pi x)) - ln(2) dx
    = 2(int from 0 to 1/2) ln(sin(2pi x)) dx - ln(2)
    = (int from 0 to 1) ln(sin(pi y)) dy - ln(2) . . . . . y = 2x, dy = 2dx
    = I - ln(2)

    I = -ln(2)

    BCHL85, where did you find this question?
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    (Original post by Jonny W)
    I
    = (int from 0 to 1) ln(sin(pi x)) dx
    = 2(int from 0 to 1/2) ln(sin(pi x)) dx
    = 2(int from 0 to 1/2) ln(cos(pi x)) dx

    2I
    = 4(int from 0 to 1/2) ln(sin(pi x)) + ln(cos(pi x)) dx
    = 4(int from 0 to 1/2) ln[sin(pi x)cos(pi x)] dx
    = 4(int from 0 to 1/2) ln[(1/2)sin(2pi x)] dx
    = 4(int from 0 to 1/2) ln(sin(2pi x)) dx - ln(2)
    = 2(int from 0 to 1/2) ln(sin(pi y)) dy - ln(2) . . . . . y = 2x, dy = 2dx
    = I - ln(2)

    I = -ln(2)

    BCHL85, where did you find this question?
    :cool: I read it in another forum.
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    (Original post by BCHL85)
    :cool: I read it in another forum.
    I was helped by the hints in a question from 1991 STEP Further Maths B. The last part of that question asks for (int from 0 to pi/2) x cot(x) dx. Rep for the first person who does it!
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    (Original post by Jonny W)
    I was helped by the hints in a question from 1991 STEP Further Maths B. The last part of that question asks for (int from 0 to pi/2) x cot(x) dx. Rep for the first person who does it!
    Would it involve the series expansion of xcotx?
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    (Original post by Jonny W)
    I was helped by the hints in a question from 1991 STEP Further Maths B. The last part of that question asks for (int from 0 to pi/2) x cot(x) dx. Rep for the first person who does it!
    I don't think this integral can be done by a simple substitution or even by parts. So it's best to concentrate on the series expansion.

    The series expansion of cotx is:

    1/x - x/3 - [x^3]/45 - 2[x^5]/945 - ...

    multiplying through x will get us xcot(x):

    1 - [x^2]/3 - [x^4]/45 - [x^6]/945 - [x^8]/4725 - ...

    Integrating the sequence will be:

    x - [x^3]/9 - [x^5]/225 - [x^7]/6615 - [x^9]/42525 - ...

    Putting in the limits will get us:

    [pi]/2 - [pi^3]/72 - [pi^5]/7200 - [pi^7]/846720 - [pi^9]/21772800

    Take a factor of Pi/2 out:

    [pi]/2 * (1 - [pi^2]/36 - [pi^4]/3600 - [pi^6]/423360 - [pi^8]/10886400 - ...)

    Now concentrate on the sum to infinity in the right expression. It follows that this converges to the limit ln2.

    So we have the INT (x*cotx) [0...pi/2] = pi/2 * ln2 = [pi.ln2]/2

    Euclid.
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    I think it can be done by parts using the result of the integral of ln sin x between 0 and pi/2 proved earlier (because the integral of cot x is ln sin x).

    The "tricky" bit is that you have to assume, or demonstrate, that as x ->0+,
    x(ln sin x) -> 0.

    Davros
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    Yes that would be all very good assuming one could integrate ln(sin(x)).

    However if the question was just "integrate x.cotx between 0 and pi/2" as it stands, then the series expansion would probably be the right direction to approaching it.

    This is only my opinion.

    Euclid.
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    well, my answer was based on Jonny W's post which said that this was the last step of a question which had previously asked you to evaluate the integral of ln sin x between the same limits.

    Integrating power series is great fun but (a) you still need to work out the sum at the end (and it may not be obvious what this is!); (b) it's not justified properly until you've done a bit of undergraduate analysis.

    The ln sin x integral seems to be an old favourite of Cambridge entrance examiners. You can do it by substitution and rearrangement as demonstrated by Jonny W, but it's not at all obvious that the integral exists, since ln sin x isn't defined at x = 0.

    Davros
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    I did abit different from u, Jonny_W. First use sinpix = 2sinpix/2.cospix/2, then let t = 1-x to prove Int(ln(cospix/2) = Int(ln(sin..)
    Then let u = x/2 to make the equation comes to I = ln2 +2I.
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    (Original post by davros)
    The ln sin x integral seems to be an old favourite of Cambridge entrance examiners. You can do it by substitution and rearrangement as demonstrated by Jonny W, but it's not at all obvious that the integral exists, since ln sin x isn't defined at x = 0.
    Convergence isn't too hard to prove because ln(sin x) ~= ln(x) for small x > 0 and

    (int from epsilon to 1) ln(x) dx
    = [x ln(x) - x](from epsilon to 1)
    = -epsilon ln(epsilon) - 1 + epsilon
    -> -1 as epsilon -> 0.
 
 
 
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