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riddles >> putnam exam (pure math) >> Integral with reciprocal of log x.
(Message started by: Aryabhatta on Jul 26th, 2009, 4:37pm)

Title: Integral with reciprocal of log x.
Post by Aryabhatta on Jul 26th, 2009, 4:37pm
True or False?

for p >= 0

Integral01 (xp-1)dx/log(x) = log(p+1)

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

Title: Re: Integral with reciprocal of log x.
Post by Obob on Jul 27th, 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((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.[/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 Integral01 (x-1) 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 27th, 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 27th, 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(xp-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.gifxp= (1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif1+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.gifphttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup0.gif(1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif1+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.gif1+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) + limhttp://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.gif1+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(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 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 27th, 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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