

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

Title: Re: Integral with reciprocal of log x. Post by Ronno on Jul 26^{th}, 2009, 11:54pm True. Proof by Mathematica ;D 

Title: Re: Integral with reciprocal of log x. Post by Obob on Jul 27^{th}, 2009, 8:22am [hide]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.[/hide] 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. 

Title: Re: Integral with reciprocal of log x. Post by Aryabhatta on Jul 27^{th}, 2009, 10:37am 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! 

Title: Re: Integral with reciprocal of log x. Post by Obob on Jul 27^{th}, 2009, 11:49am 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, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subepsilon.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup1.gif(x^{p}1) dx/log(x). We have G(0, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif) = 0, and interchanging interal and derivative is valid for http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif > 0, so http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifG/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/partial.gifp = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subepsilon.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup1.gifx^{p}= (1  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif^{1+p})/(1+p). Thus G(p, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif) =  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif_{p}http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup0.gif(1  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif^{1+p}) dp/(1+p) = log(p+1) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subp.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup0.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif^{1+p} dp/(1+p) for http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif > 0. Now since for fixed p > 1 the integrand is absolutely convergent, for fixed p we see that G(p,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif) is a continuous function of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif at http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif = 0. Therefore G(p, 0) = log(p+1) + lim_{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subp.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup0.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif^{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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0. 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) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub0.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup1.gif(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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 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. 

Title: Re: Integral with reciprocal of log x. Post by Aryabhatta on Jul 27^{th}, 2009, 12:19pm 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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