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Topic: Integral with reciprocal of log x. (Read 4869 times) 

Aryabhatta
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Integral with reciprocal of log x.
« on: Jul 26^{th}, 2009, 4:37pm » 
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True or False? for p >= 0 Integral_{0}^{1} (x^{p}1)dx/log(x) = log(p+1)

« Last Edit: Jul 26^{th}, 2009, 4:37pm by Aryabhatta » 
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Ronno
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Re: Integral with reciprocal of log x.
« Reply #1 on: Jul 26^{th}, 2009, 11:54pm » 
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True. Proof by Mathematica

« Last Edit: Jul 26^{th}, 2009, 11:55pm by Ronno » 
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Obob
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Re: Integral with reciprocal of log x.
« Reply #2 on: Jul 27^{th}, 2009, 8:22am » 
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Observe that the result is clearly true for p = 0. Differentiate both sides with respect to p, waving your hands a bit (or quoting some theorem) to push the derivative into the integral. d((x^{p}1)/log(x))/dp = x^{p}, and Integral_{0}^{1} x^{p} dx = 1/(p+1), which is the same answer we get by differentiating the RHS. This implies the LHS = RHS. Oh, and the restriction p >= 0 is unnecessary; p > 1 will do. Incidentally, on this page http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign it is suggested to evaluate Integral_{0}^{1} (x1) dx/log(x) by considering the problem given here and differentiating under the integral.

« Last Edit: Jul 27^{th}, 2009, 10:20am by Obob » 
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Aryabhatta
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Re: Integral with reciprocal of log x.
« Reply #3 on: Jul 27^{th}, 2009, 10:37am » 
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Correct! Not sure about the p > 1. The theorem statement I had at hand (in Apostol's analysis book) seemed to require df(x,p)/dp to be bounded for the differentiation under the integral sign to work. Do you know of any other version of the theorem which lets p > 1? The wiki page having this is just a coincidence!


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Obob
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Re: Integral with reciprocal of log x.
« Reply #4 on: Jul 27^{th}, 2009, 11:49am » 
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The result might not be quite as straightforward, but it should certainly be true. One way to set it up would be to look instead at G(p, ) = (x^{p}1) dx/log(x). We have G(0, ) = 0, and interchanging interal and derivative is valid for > 0, so G/p = x^{p}= (1  ^{1+p})/(1+p). Thus G(p, ) =  _{p}(1  ^{1+p}) dp/(1+p) = log(p+1) + ^{1+p} dp/(1+p) for > 0. Now since for fixed p > 1 the integrand is absolutely convergent, for fixed p we see that G(p,) is a continuous function of at = 0. Therefore G(p, 0) = log(p+1) + lim_{0} ^{1+p} dp/(1+p). But the limit on the RHS is clearly zero for p > 1 by the dominated convergence theorem since the integrand is dominated by a bounded function and approaches 0 pointwise almost everywhere as 0. Another big hammer approach which should be valid: allow p to be complex. If the real part of p is bigger than 1, then F(p) = (x^{p}1) dx/log(x) is defined. Moreover, I think it should be possible to show F is a complex analytic function. But it agrees with log(p+1) for Re p 0, which implies it must equal log(p+1) everywhere it is defined. Generally speaking, I think absolute convergence is the main thing you have to be worried about when applying this kind of argument. There is another argument here http://mathworld.wolfram.com/IntegrationUndertheIntegralSign.html which derives the same formula by applying Fubini's theorem (interchanging two integrals). The only thing you really have to worry about for Fubini's theorem is absolute convergence.

« Last Edit: Jul 27^{th}, 2009, 12:07pm by Obob » 
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Aryabhatta
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Re: Integral with reciprocal of log x.
« Reply #5 on: Jul 27^{th}, 2009, 12:19pm » 
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Right, I was expecting some arguments like that to work... but was looking at directly applying a stronger theorem, seems like Fubini's is one which will work. Thanks!

« Last Edit: Jul 27^{th}, 2009, 12:19pm by Aryabhatta » 
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