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Math Help - Integration by parts using known limit

  1. #1
    Member courteous's Avatar
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    Question Integration by parts using known limit

    I have a sum \displaystyle{  \sum_{k=0}^\infty \frac{1}{k!}\left\{\int_0^1 (x\ln x)^k dx\right\}  }.

    Now, per partes it "is easy to show that" \displaystyle{  \int_0^1 (x\ln x)^k dx=\frac{(-1)^kk!}{(k+1)^{k+1}}  } (for this I should use \displaystyle{  \lim_{x\to 0}x\ln x=0   }).

    I "know" that the \lim holds ... but how to use it in per partes integration?
    Last edited by courteous; November 7th 2010 at 11:02 PM. Reason: 'x' goes to 0 (not INFINITY)
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    the limit does not hold, it's obviously false, can't you just see it?

    as for the integral put t=-\ln x and invoke the gamma function.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by courteous View Post
    I have a sum \displaystyle{  \sum_{k=0}^\infty \frac{1}{k!}\left\{\int_0^1 (x\ln x)^k dx\right\}  }.

    Now, per partes it "is easy to show that" \displaystyle{  \int_0^1 (x\ln x)^k dx=\frac{(-1)^kk!}{(k+1)^{k+1}}  } (for this I should use \displaystyle{  \lim_{x\to \infty}x\ln x=0   }).

    I "know" that the \lim holds ... but how to use it in per partes integration?
    Let consider first the indefinite integral...

    \displaystyle \int t^{m}\ \ln ^{n} t\ dt (1)

    ... with m and n non negative integers. Integrating by parts we obtain...

    \displaystyle \int t^{m}\ \ln ^{n} t\ dt = \frac{t^{m+1}}{m+1}\ \ln^{n} t - \frac{n}{m+1}\ \int t^{m}\ \ln^{n-1} t\ dt (2)

    The integral in (2) is similar to the integral (1), but the exponent of log is n-1 instead of n. That suggests to integrate n times by part so that we obtain...

    \displaystyle \int t^{m}\ \ln ^{n} t\ dt = t^{m+1}\ \sum_{i=0}^{n} \frac{(-1)^{i}}{(m+1)^{i+1}}\ \frac{n!}{(n-i)!}\ \ln ^{n-i} t +c (3)

    The second step is the computation of...

    \displaystyle \lim_{t \rightarrow 0} t^{m}\ \ln^{n} t (4)

    ... and that is done applying n times l'Hopital's rule...

    \displaystyle \lim_{t \rightarrow 0} t^{m}\ \ln^{n} t = \lim_{t \rightarrow 0} (-1)^{n} \frac{n!}{m^{n}}\ t^{m} = 0 for m>0 (5)

    Combining (3) and (5) with some easy steps we arrive to write...

    \displaystyle \int_{0}^{1}t^{m}\ \ln^{n} t\ dt = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}} (6)

    ... that for m=n=k becomes...

    \displaystyle \int_{0}^{1}t^{k}\ \ln^{k} t\ dt = (-1)^{k}\ \frac{k!}{(k+1)^{k+1}} (7)

    Kind regards

    \chi \sigma
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