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Topic: Integral with reciprocal of log x. (Read 4879 times) |
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Aryabhatta
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Integral with reciprocal of log x.
« on: Jul 26th, 2009, 4:37pm » |
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True or False?
for p >= 0
Integral01 (xp-1)dx/log(x) = log(p+1)
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| « Last Edit: Jul 26th, 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 26th, 2009, 11:54pm » |
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True.
Proof by Mathematica 
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| « Last Edit: Jul 26th, 2009, 11:55pm by Ronno » |
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Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
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Obob
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Re: Integral with reciprocal of log x.
« Reply #2 on: Jul 27th, 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((xp-1)/log(x))/dp = xp, and Integral01 xp 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 Integral01 (x-1) dx/log(x) by considering the problem given here and differentiating under the integral.
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| « Last Edit: Jul 27th, 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 27th, 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 27th, 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,  ) =    (xp-1) dx/log(x). We have G(0, ) = 0, and interchanging interal and derivative is valid for > 0, so
 G/p = xp= (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) =  (xp-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.
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| « Last Edit: Jul 27th, 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 27th, 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!
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| « Last Edit: Jul 27th, 2009, 12:19pm by Aryabhatta » |
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