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Math Help - Hint at solving an integral, please?

  1. #1
    Member Pranas's Avatar
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    Hint at solving an integral, please?

    Hello.

    \displaystyle \[I(p) = \int\limits_0^{ + \infty } {\frac{{{x^{p - 1}} \cdot \ln (x)}}{{x + 1}}dx} \]

    and

    \displaystyle \[\frac{d}{{dp}}I(p) = \int\limits_0^{ + \infty } {\frac{{{x^{p - 1}} \cdot {{(\ln (x))}^2}}}{{x + 1}}dx} \]

    seems useless to me.
    So I tried substituting

    \displaystyle \[\ln (x + 1) = u\]

    which apparently gave me

    \displaystyle \[I(p) = \int\limits_0^{ + \infty } {{e^{(u - 1)(p - 1)}}(u - 1)du} \]

    I am not sure if this is really true. Moreover, I don't seem to be able to solve them (I actually expect to get an Euler's gamma function somehow).
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  2. #2
    Junior Member
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    Try this way:

    u=(x^(p-1))/(x+1) and dv=ln(x)dx. Find du and v, then you will get I(p)=u*v-Integrate[(v*du)]
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  3. #3
    Member Pranas's Avatar
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    Quote Originally Posted by derdack View Post
    Try this way:

    u=(x^(p-1))/(x+1) and dv=ln(x)dx. Find du and v, then you will get I(p)=u*v-Integrate[(v*du)]
    I get nothing elegant coming out of it, just more complicated problems.

    Answer is supposed to be \displaystyle \[\frac{{ - {\pi ^2} \cdot \cos (\pi p)}}{{{{(\sin (\pi p))}^2}}}\]

    which leaves me almost certain this can be achieved by making Euler's beta function


    Edit: It took some hours of pure thinking, but satisfaction is well worth it.

    \displaystyle \[I(p) = \int\limits_0^{ + \infty } {\frac{{{x^{p - 1}}\cdot\ln (x)}}{{x + 1}}dx}  = \frac{d}{{dp}}\int\limits_0^{ + \infty } {\left( {\int {\frac{{{x^{p - 1}}\cdot\ln (x)}}{{x + 1}}dp} } \right)dx}  = \frac{d}{{dp}}\int\limits_0^{ + \infty } {\frac{{{x^{p - 1}}}}{{x + 1}}dx} \]

    Integral is already a form of Euler's beta function and known manipulations can be done:

    \displaystyle \[\int\limits_0^{ + \infty } {\frac{{{x^{p - 1}}}}{{x + 1}}dx}  = {\rm B}\left( {p;1 - p} \right) = \frac{\pi }{{\sin (\pi p)}}\]

    Hence

    \displaystyle \[I(p) = \frac{d}{{dp}}\frac{\pi }{{\sin (\pi p)}} = \frac{{ - {\pi ^2} \cdot \cos (\pi p)}}{{{{(\sin (\pi p))}^2}}}\]
    Last edited by Pranas; April 7th 2011 at 01:03 PM.
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