wu :: forums « wu :: forums - Integral with reciprocal of log x. » Welcome, Guest. Please Login or Register. Sep 22nd, 2023, 11:42pm RIDDLES SITE WRITE MATH! Home Help Search Members Login Register
 wu :: forums    riddles    putnam exam (pure math) (Moderators: towr, Icarus, Grimbal, Eigenray, william wu, SMQ)    Integral with reciprocal of log x. « Previous topic | Next topic »
 Pages: 1 Reply Notify of replies Send Topic Print
 Author Topic: Integral with reciprocal of log x.  (Read 4856 times)
Aryabhatta
Uberpuzzler

Gender:
Posts: 1321
 Integral with reciprocal of log x.   « on: Jul 26th, 2009, 4:37pm » Quote Modify

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
Junior Member

Gender:
Posts: 140
 Re: Integral with reciprocal of log x.   « Reply #1 on: Jul 26th, 2009, 11:54pm » Quote Modify

True.
Proof by Mathematica
 « Last Edit: Jul 26th, 2009, 11:55pm by Ronno » IP Logged

Sarchasm: The gulf between the author of sarcastic wit and the person who doesn't get it..
Obob
Senior Riddler

Gender:
Posts: 489
 Re: Integral with reciprocal of log x.   « Reply #2 on: Jul 27th, 2009, 8:22am » Quote Modify

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
Uberpuzzler

Gender:
Posts: 1321
 Re: Integral with reciprocal of log x.   « Reply #3 on: Jul 27th, 2009, 10:37am » Quote Modify

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!
 IP Logged
Obob
Senior Riddler

Gender:
Posts: 489
 Re: Integral with reciprocal of log x.   « Reply #4 on: Jul 27th, 2009, 11:49am » Quote Modify

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
Uberpuzzler

Gender:
Posts: 1321
 Re: Integral with reciprocal of log x.   « Reply #5 on: Jul 27th, 2009, 12:19pm » Quote Modify

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
 Pages: 1 Reply Notify of replies Send Topic Print

 Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs => putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »