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Math Help - Integral with logarithm

  1. #1
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    Krizalid's Avatar
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    Integral with logarithm

    Compute \int_0^\infty\frac{\ln\big(1+x^2\big)}{x^{1+\alpha  }}\,dx for 0<\alpha<2.
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  2. #2
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    Quote Originally Posted by Krizalid View Post
    Compute \int_0^\infty\frac{\ln\big(1+x^2\big)}{x^{1+\alpha  }}\,dx for 0<\alpha<2.
    First integrate by parts to get with
    u=\ln(x^2+1) \implies du=\frac{2x}{x^2+1} and
    dv=x^{-(\alpha+1)} \implies v=-\frac{1}{\alpha}x^{-\alpha}

    This Formally gives

    \int_0^\infty\frac{\ln\big(1+x^2\big)}{x^{1+\alpha  }}\,dx=-\frac{1}{\alpha}x^{-\alpha}\ln(x^2+1) \bigg|_{0}^{\infty}+\int_{0}^{\infty}\frac{1}{\alp  ha}x^{-\alpha}\frac{2x}{x^2+1}dx

    Now justifying the limits

    \lim_{x\to \infty}\frac{\ln(x^2+1)}{x^\alpha}=\lim_{x \to \infty}\frac{\frac{2x}{x^2+1}}{\alpha x^{\alpha-1}} =\lim_{x \to \infty}\frac{2x^2}{(x^2+1)}\frac{1}{x^\alpha }=1\cdot 0

    \lim_{x\to 0}\frac{\ln(x^2+1)}{x^\alpha}=\lim_{x \to 0}\frac{\frac{2x}{x^2+1}}{\alpha x^{\alpha-1}} =\lim_{x\to 0}\frac{2x^{2-\alpha}}{1}\frac{1}{ (x^2+1)}=0 \cdot 1=0

    Both of the above limits only exsist when 0 < \alpha < 2

    This gives

    \int_0^\infty\frac{\ln\big(1+x^2\big)}{x^{1+\alpha  }}\,dx=\frac{1}{\alpha}\int_{0}^{\infty}x^{-\alpha}\frac{2x}{x^2+1}dx

    Now let t=x^2 \implies t^{\frac{1}{2}}=x \implies dt=2xdx

    This turns the right hand side into a form of the Beta function!

    \frac{1}{\alpha}\int_{0}^{\infty}x^{-\alpha}\frac{2x}{x^2+1}dx=\frac{1}{\alpha}\int_{0}  ^{\infty}\frac{t^{-\frac{\alpha}{2}}}{(t+1)}dt=\frac{1}{\alpha}\beta \left( \frac{2-\alpha}{2},\frac{\alpha}{2}\right)=\frac{1}{\alpha  }\Gamma\left( \frac{2-\alpha}{2}\right)\Gamma\left( \frac{\alpha}{2}\right)

    Now using Euler reflection formula the product of the two Gamma functions can be written as

    \frac{1}{\alpha}\Gamma\left( \frac{2-\alpha}{2}\right)\Gamma\left( \frac{\alpha}{2}\right)=\frac{1}{\alpha}\Gamma\lef  t( 1-\frac{\alpha}{2}\right)\Gamma\left( \frac{\alpha}{2}\right) =\frac{\pi}{\alpha \sin\left( \frac{\pi \alpha}{2}\right)}
    Last edited by TheEmptySet; March 5th 2010 at 08:46 PM. Reason: add part about Euler reflection formula
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  3. #3
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    Are you expecting a beautiful solution ?

    Intuitively , I think there must be another method which doesn't require any knowledges of Gamma function .
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  4. #4
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    no, i'm not looking for a beautiful solution, i just wanted to share this problem.

    here's my solution:

    \int_{0}^{\infty }{\frac{x^{1-\alpha}}{1+x^{2}t}\,dx}=\frac{1}{2}t^{\frac{\alpha  }{2}-1}\int_{0}^{\infty }{\frac{x^{-\frac{\alpha}{2}}}{1+x}\,dx}=\frac{1}{2}t^{\frac{\  alpha}{2}-1}\beta \left( 1-\frac{\alpha}{2},\frac{\alpha}{2} \right), thus the integral equals \frac{1}{2}t^{\frac{\alpha}{2}-1}\Gamma \left( 1-\frac{\alpha}{2} \right)\Gamma \left( \frac{\alpha}{2} \right)=\frac{\pi }{2}t^{\frac{\alpha}{2}-1}\csc \left( \frac{\alpha\pi }{2} \right).

    hence \int_{0}^{\infty }{\frac{\ln \left( 1+x^{2} \right)}{x^{1+\alpha }}}\,dx=\int_{0}^{\infty }{\int_{0}^{1}{\frac{x^{1-\alpha }}{1+x^{2}t}\,dt}\,dx}=\frac{\pi }{2}\csc \left( \frac{a\pi }{2} \right)\int_{0}^{1}{t^{\frac{a}{2}-1}\,dt}=\frac{\pi }{a}\csc \left( \frac{a\pi }{2} \right).
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