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Math Help - more integrals

  1. #1
    Super Member Random Variable's Avatar
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    more integrals

    Don't hide solutions.


    1)  \int_{0}^{\frac{\pi}{2}}} \frac{\arctan(\sin x)}{\sin x} \ dx


    2)  \int^{\infty}_{0} \frac{\sin x \ln x}{x} \ dx


    3)  \int_{0}^{\frac{\pi}{2}}} \ln(\cos x) \ln(\sin x) \cos x \sin x \ dx
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  2. #2
    Super Member Random Variable's Avatar
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    Re: more integrals

    Please don't give hints.
    Last edited by Random Variable; October 16th 2011 at 01:55 PM.
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    Re: more integrals

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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: more integrals

    We have \lim_{x\to 0^+}\frac{\arctan(\lambda \sin x)}{\sin x}=\ldots=\lambda , so the following function is well defined (by means of a convergent integral):

    I:[0,+\infty)\to \mathbb{R},\;\;I(\lambda)=\int_0^{\pi/2}\frac{\arctan (\lambda \sin x)}{\sin x}dx

    Then I'(\lambda)=\int_0^{\pi/2}\frac{dx}{1+\lambda^2\sin^2 x}

    Using the substitution t=\tan x:

    I'(\lambda)=\int_0^{+\infty}\frac{ \frac{dt}{1+t^2} }{ 1+\frac{\lambda^2 t^2}{1+t^2} }=\int_0^{+\infty}\frac{dt}{(1+\lambda^2)t^2+1}

    Using the substitution u=\sqrt{1+\lambda^2}t:

    I'(\lambda)=\frac{1}{\sqrt{1+\lambda^2}}\int_{0}^{  +\infty}\frac{du}{u^2+1}=\frac{1}{\sqrt{1+\lambda^  2}}\dfrac{\pi}{2}

    Integrating both sides:

    I(\lambda)=\dfrac{\pi}{2}\log |\lambda+\sqrt{\lambda^2+1}|+C

    For \lambda=0 we have 0=0+C that is, C=0. Then, I(\lambda)=\dfrac{\pi}{2}\log |\lambda+\sqrt{\lambda^2+1}|.

    As a consequence:

    \int_0^{\pi/2}\frac{\arctan (\sin x)}{\sin x}dx=I(1)=\dfrac{\pi}{2}\log (1+\sqrt{2})=\boxed{\dfrac{\pi}{2}\sinh^{-1}(1)}
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  5. #5
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    Re: more integrals

    Consider  ab = \frac{1}{2}[a^2 + b^2 - (a-b)^2] , we have

     \ln[\sin(x)]\ln[\cos(x)] = \frac{1}{2} \left( \ln^2[\sin(x)] + \ln^2[\cos(x)] - \ln^2[\tan(x)] \right)

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

    = \int_0^{\frac{\pi}{2}}\ln^2[\cos(x)]\sin(x)\cos(x) ~dx = \int_0^1 x\ln^2(x)~dx = \frac{1}{4}

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

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

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

     = \left[ \frac{x^2\ln^2(x)}{x^2+ 1}\right]_0^1 - 2\int_0^1 \frac{x\ln(x)}{x^2+1}~dx = \frac{\pi^2}{24}

    so the answer is  \frac{1}{4} - \frac{\pi^2}{48}
    Last edited by simplependulum; December 5th 2011 at 12:35 AM.
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  6. #6
    Member sbhatnagar's Avatar
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    Re: more integrals

    Quote Originally Posted by uniquesailor View Post
    Can you explain what is 'MAZ identity'?
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