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        riddles >> medium >> Logarithmic integral
        
(Message started by: pex on Jul 12^th, 2008, 12:44pm)

Title: Logarithmic integral
Post by pex on Jul 12^th, 2008, 12:44pm

If n is a nonnegative integer, calculate the integral

1
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif (1 - ln x)ⁿ dx.
0

Title: Re: Logarithmic integral
Post by Barukh on Jul 13^th, 2008, 11:56am

[hideb]
Denote the sought integral I(n). Make a substitution x = e^y, 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! http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_{k = 0...n} 1/k!

Is there a better formula? :-/

Note that I(n)/n! converges to e.
[/hideb]

Title: Re: Logarithmic integral
Post by towr on Jul 13^th, 2008, 12:39pm

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.
[hide](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⁾ |_inf^e = inf ?!
[/hide] :-/

Title: Re: Logarithmic integral
Post by pex on Jul 13^th, 2008, 1:06pm

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 ;)

Title: Re: Logarithmic integral
Post by towr on Jul 13^th, 2008, 1:10pm

on 07/13/08 at 13:06:43, pex wrote:

@towr: d(1/y) is not the same as (1/y)dy ;)

Doh. Thanks.

Quote:

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).

Title: Re: Logarithmic integral
Post by pex on Jul 13^th, 2008, 1:25pm

on 07/13/08 at 13:10:55, towr wrote:

The sequence this integral give is: "The number of one-to-one sequences that can be formed from n distinct objects."

Well, yes, after all it's [hide]sum_k=0..n ( n! / k! ) = sum_k=0..n (number of permutations of k objects chosen from n) = (number of permutations of any number of objects chosen from n)[/hide], so that makes sense.

on 07/13/08 at 13:10:55, 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."

Hmm... so I(n) = e*Gamma(n+1,1) = e*(n! - int_x=0..1xⁿe^-xdx). It's interesting, but I'm not sure how much it helps...

Title: Re: Logarithmic integral
Post by pex on Jul 13^th, 2008, 1:44pm

on 07/13/08 at 13:10:55, 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."

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 [hide]did the integration by parts immediately, obtaining a second-order recurrence relation[/hide] instead.)