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pex
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Logarithmic integral
« on: Jul 12th, 2008, 12:44pm » |
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If n is a nonnegative integer, calculate the integral
1
 (1 - ln x)n dx.
0
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Barukh
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Re: Logarithmic integral
« Reply #1 on: Jul 13th, 2008, 11:56am » |
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Denote the sought integral I(n). Make a substitution x = ey, integrate by parts to get the recursive relation
I(n) = nI(n-1) + 1
Note that I(0) = 1. Using induction, we can arrive at the following formula:
I(n) = n!  k = 0...n 1/k!
Is there a better formula? 
Note that I(n)/n! converges to e.
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towr
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Re: Logarithmic integral
« Reply #2 on: Jul 13th, 2008, 12:39pm » |
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There's definitely something wrong with what I've done; but i fear I haven't done enough integration and stuff lately to know what anymore.
(1 - lnx)n dx = (ln e/x)n dx
substitute x=e/y
(ln y)n d (e/y) = e (ln y)n d (1/y) = e/(n+1) d (ln y)(n+1)
integration limits, x:0->1 => y:inf->e
e/(n+1) (ln y)(n+1) |infe = inf ?!
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| « Last Edit: Jul 13th, 2008, 12:44pm by towr » |
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pex
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Re: Logarithmic integral
« Reply #3 on: Jul 13th, 2008, 1:06pm » |
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Barukh's answer (the summation formula) is what I had, including the observation on the asymptotic behavior. I also haven't been able to find a more "closed-form" type of expression. But, given what the answer looks like: is there any interpretation to this result, or is it just a coincidence? (I don't know, but I'm afraid it's the latter.)
@towr: d(1/y) is not the same as (1/y)dy 
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towr
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Re: Logarithmic integral
« Reply #4 on: Jul 13th, 2008, 1:10pm » |
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on Jul 13th, 2008, 1:06pm, pex wrote:| @towr: d(1/y) is not the same as (1/y)dy |
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Doh. Thanks.
Quote:| Barukh's answer (the summation formula) is what I had, including the observation on the asymptotic behavior. I also haven't been able to find a more "closed-form" type of expression. But, given what the answer looks like: is there any interpretation to this result, or is it just a coincidence? (I don't know, but I'm afraid it's the latter.) |
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http://www.research.att.com/~njas/sequences/A000522
The sequence this integral give is: "The number of one-to-one sequences that can be formed from n distinct objects."
And the result can also be given as "a(n)=exp(1)*Gamma(n+1,1) where Gamma(z,t)=Integral_{x>=t} exp(-x)x^(z-1) dx is incomplete gamma function." (Something quickmath eventually helped me find as well, although I hadn't a clue what it meant with gamma(n+1, 1); I only knew the normal gamma function).
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| « Last Edit: Jul 13th, 2008, 1:11pm by towr » |
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pex
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Re: Logarithmic integral
« Reply #5 on: Jul 13th, 2008, 1:25pm » |
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on Jul 13th, 2008, 1:10pm, towr wrote:| The sequence this integral give is: "The number of one-to-one sequences that can be formed from n distinct objects." |
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Well, yes, after all it's sumk=0..n ( n! / k! ) = sumk=0..n (number of permutations of k objects chosen from n) = (number of permutations of any number of objects chosen from n), so that makes sense.
on Jul 13th, 2008, 1:10pm, towr wrote:| And the result can also be given as "a(n)=exp(1)*Gamma(n+1,1) where Gamma(z,t)=Integral_{x>=t} exp(-x)x^(z-1) dx is incomplete gamma function." |
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Hmm... so I(n) = e*Gamma(n+1,1) = e*(n! - intx=0..1xne-xdx). It's interesting, but I'm not sure how much it helps...
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pex
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Re: Logarithmic integral
« Reply #6 on: Jul 13th, 2008, 1:44pm » |
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on Jul 13th, 2008, 1:10pm, towr wrote:| And the result can also be given as "a(n)=exp(1)*Gamma(n+1,1) where Gamma(z,t)=Integral_{x>=t} exp(-x)x^(z-1) dx is incomplete gamma function." |
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Actually, with (a slight variation on) Barukh's substitution, this is pretty obvious: substitute y = 1 - ln x. (I didn't immediately notice it, because I hadn't used such a substitution; I did the integration by parts immediately, obtaining a second-order recurrence relation instead.)
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