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Math Help - Compute an integral

  1. #1
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    Compute an integral

    Find \int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx.
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  2. #2
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    Quote Originally Posted by Krizalid View Post
    Find \int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx.
    substitute x = \sin^2t to get \int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx=4 \int_0^{\frac{\pi}{2}} \ln(\sin t) \ dt = -2\pi \ln 2.
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  3. #3
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    yes, but can you find another way to solve it?
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    Another way? What does (s)he want? Hmm ...

    Let I=\int_0^1\frac{\ln x}{\sqrt{x-x^2}}\mathrm dx and let J=\int_0^1\frac{\ln x}{\sqrt{1-x^2}}\mathrm dx.

    The substitution x=u^2 shows that I=4J.

    The substituion y=1-x shows that I=\int_0^1\frac{\ln (1-x)}{\sqrt{x-x^2}}\mathrm dx and so 2I=\int_0^1\frac{\ln x+\ln(1-x)}{\sqrt{x-x^2}}\mathrm dx=\int_0^1\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx.

    Thus I=\frac12\int_0^1\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx<br />
=\int_0^{1/2}\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx by symmetry.

    Now put u=2\sqrt{x-x^2} so that 1-u^2=(1-2x)^2, 1-2x=\sqrt{1-u^2} for 0\leq x\leq{\textstyle\frac12} and 0\leq u\leq 1, and \mathrm dx=\frac{u\,\mathrm du}{2\sqrt{1-u^2}}.

    Thus I=\int_0^1\frac{\ln (u^2/4)}{u/2}\frac{u\,\mathrm du}{2\sqrt{1-u^2}}=\int_0^1\frac{2\ln u-\ln 4}{\sqrt{1-u^2}}\mathrm du=2J-\ln 4\bigl[\arcsin u\bigr]_0^1={\textstyle\frac12}I-\pi\ln 2.

    Thus I=-2\pi\ln 2.

    Wait a moment, this is merely a disguised version of the \int_0^{\pi/2}\ln(\sin x)\,\mathrm dx method. It's no better than NonCommAlg's rejected solution.

    (Thinks: Perhaps I should have used an infinite series, or some kind of limit, or introduced a parameter, or ... (Mind boggles at this point))

    An answer that is not really an answer? This is trickier than it looks.

    Hmm ...
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  5. #5
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    Quote Originally Posted by halbard View Post

    Wait a moment, this is merely a disguised version of the \int_0^{\pi/2}\ln(\sin x)\,\mathrm dx method. It's no better than NonCommAlg's rejected solution.
    I never rejected NCA's solution, it's just that the integral is quite trivial by using the fact you stated.

    It's a well-known fact and easy to prove, so that's why he left details omitted.

    I have another solution, I'll post it later, but I like yours.
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  6. #6
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    What about this ?

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

     = \lim_{a\to0} \frac{1}{a} [ \int_0^1 \frac{x^a-1}{\sqrt{x-x^2}}~dx ]


     = \lim_{a\to0} \frac{1}{a} \sqrt{\pi} [ \frac{ \Gamma(a + \frac{1}{2})}{\Gamma(a + 1) } - \sqrt{\pi} ]

    From the Lengre's Identity

     = \lim_{a\to0} \frac{1}{a} \sqrt{\pi} [ \frac{\sqrt{\pi} \Gamma(2a) 2^{-2a+1}}{ \Gamma(a) \Gamma(a + 1) } - \sqrt{\pi} ]

     = \pi \lim_{a\to0} \frac{1}{a}[ 2^{-2a+1} / 2 - 1]

     = \pi \lim_{a\to0} \frac{1}{a}[ 2^{-2a} - 1]

     = \pi \ln(2) (-2) = - 2 \pi \ln(2)


    I think it is the best method by substituting  x = \sin^2{\theta} ..
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