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Math Help - Is this integral correct?

  1. #1
    Super Member Random Variable's Avatar
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    Is this integral correct?

    I've been working on this for the past hour.


     \int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx \ (a>0)

    let  u = \frac{a}{x}

    then  du = -\frac{a}{x^{2}}dx = -\frac{u^{2}}{a} dx

     \int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx = -\int_{\infty}^{0} \frac{\ln\Big( \frac{a}{u}\Big)}{a^{2}+ \frac{a^{2}}{u^{2}}} \ \frac{a}{u^{2}} du =  \frac{1}{a} \int_{0}^{\infty} \frac{\ln\Big( \frac{a}{u}\Big)}{u^{2}+1} \ du


     = \frac{\ln a}{a} \int_{0}^{\infty} \frac{1}{u^{2}+1} \ du -\frac{1}{a} \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du



     \int_{0}^{\infty} \frac{1}{u^{2}+1} \ du = \tan^{-1} (\infty) - \tan^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}



     \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du

    let  w = \frac{1}{u}

    then  dw = -\frac{1}{u^{2}} du = - w^{2} du

     \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du = -\int_{\infty}^{0} \frac{\ln \Big( \frac{1}{w}\Big)}{\frac{1}{w^{2}}+1} \ \frac{1}{w^{2}} \ dw  = -\int_{0}^{\infty} \frac{\ln w}{w^{2}+1} \ dw


    which implies that  \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du =0



    so finally  \int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx = \frac{\ln a}{a} \Big(\frac{\pi}{2}\Big) -\frac{1}{a} \Big(0\Big) = \frac{\pi \ln a}{2a} ?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Its correct according to Mathcad
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Calculus26 View Post
    Its correct according to Mathcad
    Thanks. It seemed to be correct for at least a few different values of a.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    Mathcad does the symbolic calculation for arbitrary a-- its ruining my skills
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Do you know the residue theorem? I haven't checked fully but it seems that such an integral would yield fairly easily to it.
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  6. #6
    Moo
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    Hello,

    As Bruno J. mentioned, it can be done with the residue theorem.

    If you consider the keyhole contour, then it can be proved that :

    \int_0^\infty R(x)\ln(x) ~dx=-\frac 12 ~ \Re \left(\sum \text{Res}\left\{R(z)(\log(z))^2\right\}\right)

    where the branch of the logarithm is such that -\pi<\text{arg}(z)\leq \pi

    It's not more beautiful than your method though.
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  7. #7
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Do you know the residue theorem? I haven't checked fully but it seems that such an integral would yield fairly easily to it.
    I didn't even think about using contour integration. The problem is actually an exercise in my complex analysis textbook.

    Moo suggested using the keyhole contour. My book suggests using a semicircle in the upper-half plane with a little semicircle about the origin (because ln z is undefined for z=0). They also suggest using the branch  \log_{-\frac{\pi} {2}} = ln|z| + i \theta \ (-\frac{\pi}{2} < \theta \le \frac{3 \pi}{2}) so that ln z is analytic on and inside of the contour.


    So let R be the radius of the big semicircle and r be the radius of the small semicircle

    and let  f(z) = \frac{\log_{\frac{-\pi}{2}} z}{a^{2}+z^{2}}

    then  \int_{C} f(z) dz = \int^{-r}_{-R} f(x) dx + \int_{-C_{r}} f(z) dz + \int^{R}_{r} f(x) dx + \int_{C_{R}} f(z) dz

     = \int^{-r}_{-R} \frac{ \ln |x| + i \pi}{a^{2}+z^{2}} \ dx + \int_{-C_{r}} f(z) dz + \int^{R}_{r} \frac {\ln x}{a^{2} + x^{2}} \ dx + \int_{C_{R}} f(z) dz


     \int_{C} f(z) dz = \int_{C} \frac{\log_{-\frac{\pi}{2}} z} {a^{2}+z^{2}} \ dz = \int_{C} \frac{\log_{-\frac{\pi}{2}} z} {(z-ai)(z+ai)} \ dz

    so there is a simple pole inside of the contour at  z_{0} = ai

     \text{Res} \{f,ai\} = \lim_{z \to ai} \frac{\log_{-\frac{\pi}{2}} z}{z+ai} = \frac{\ln|ai| + i\frac{\pi}{2}}{2ai} = \frac{\ln a + i\frac{\pi}{2}}{2ai}

    then  \int_{C} f(z) dz = 2 \pi i \ \text{Res} \{f, ai\} = \frac{\pi \ln a}{a} + i \frac{\pi^{2}}{2a}


    So I have  \int^{-r}_{-R} \frac{ \ln |x| + i \pi}{a^{2}+z^{2}} \ dx + \int_{-C_{r}} f(z) dz + \int^{R}_{r} \frac {\ln x}{a^{2} + x^{2}} \ dx + \int_{C_{R}} f(z) dz = \frac{\pi \ln a}{a} + i \frac{\pi^{2}}{2a}


    now letting R go to infinity and r go to zero

     \int^{0}_{-\infty} \frac{ \ln |x| + i \pi}{a^{2}+z^{2}} \ dx + 0 + \int^{\infty}_{0} \frac {\ln x}{a^{2} + x^{2}} \ dx + 0 = \frac{\pi \ln a}{a} + i \frac{\pi^{2}}{2a}

    I'm just assuming that the two contour integrals evaluate to zero. I'll try to offer a justification in another post. But I'll probably need some help.


    Equating real parts on both sides I have

     \int^{0}_{-\infty} \frac{ \ln |x|}{a^{2}+z^{2}} \ dx + \int^{\infty}_{0} \frac {\ln x}{a^{2} + x^{2}} \ dx = \int^{\infty}_{0} \frac{ \ln x }{a^{2}+z^{2}} \ dx + \int^{\infty}_{0} \frac {\ln x}{a^{2} + x^{2}} \ dx = \frac{\pi \ln a}{a}

    so finally  \int^{\infty}_{0} \frac {\ln x}{a^{2} + x^{2}} \ dx = \frac{\pi \ln a}{2a}

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  8. #8
    Super Member Random Variable's Avatar
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    nothing to see here
    Last edited by Random Variable; August 20th 2009 at 03:47 PM.
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  9. #9
    Super Member Random Variable's Avatar
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    OK. After a bit of confusion about the reverse triangle inequality, I think I can show that

     \lim_{R \to \infty} \int_{C_{R}} f(z) dz = \lim_{r \to 0}\int_{-C_{r}} f(z) dz = 0


     \Big|\int_{C_{R}} f(z) dz \Big|= \Big|\int_{C_{R}} \frac{\log_{-\frac{\pi}{2}}z}{a^{2}+z^{2}} \ dz\Big|


    let  z = Re^{i \theta}

    then  dz = iRe^{i \theta} d \theta


     = \Big|\int_{0}^{\pi}\frac{\ln|Re^{i \theta}| + i \theta}{a^{2} + (Re^{i \theta})^{2}} iRe^{i \theta} \ d \theta \Big|  = \int^{\pi}_{0} \Big|\frac{\ln R + i \theta}{a^{2} + (Re^{i \theta})^{2}}\Big| \Big|iRe^{i \theta}\Big| \ d \theta


     \int^{\pi}_{0} \Big|\frac{\ln R + i \theta}{a^{2} + (Re^{i \theta})^{2}}\Big| R \ d \theta \le \int^{\pi}_{0} \frac{R \ln R + R\theta}{|a^{2} + (Re^{i \theta})^{2}|} \ d \theta \le \int^{\pi}_{0} \frac{R \ln R + R \theta}{-a^{2} + R^{2}} \ d \theta \ \ (\text{since} \ R >a)


     = \frac{R \ln R}{-a^{2} + R^{2}} \int_{0}^{\pi} d \theta + \frac{R}{-a^{2}+R^{2}}\int_{0}^{\pi} \theta \ d \theta


     = \frac{2\pi R\ln R + \pi^{2}}{-2(a^{2}+R^{2})}


     \lim_{R \to \infty} \frac{2\pi R\ln R + \pi^{2}}{2(-a^{2}+R^{2})} = \lim_{R \to \infty} \frac{2 \pi \ln R + 2 \pi}{4R} = 0

    which according to the ML inequality shows that \lim_{R \to \infty}\int_{C_{R}} f(z) dz = 0


    The other one is similar except r<a.
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