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   Integral with reciprocal of log x.
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   Author  Topic: Integral with reciprocal of log x.  (Read 4789 times)
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)
« Last Edit: Jul 26th, 2009, 4:37pm by Aryabhatta » IP Logged
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 Grin
« Last Edit: Jul 26th, 2009, 11:55pm by Ronno » IP Logged

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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.
« Last Edit: Jul 27th, 2009, 10:20am by Obob » IP Logged
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) + lim0 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.
« Last Edit: Jul 27th, 2009, 12:07pm by Obob » IP Logged
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!
« Last Edit: Jul 27th, 2009, 12:19pm by Aryabhatta » IP Logged
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