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Re: Making a function out of repeated hyperbola integrations?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122078] Re: Making a function out of repeated hyperbola integrations?
  • From: Jacopo Bertolotti <jacopo.bertolotti at gmail.com>
  • Date: Wed, 12 Oct 2011 03:43:56 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201110110824.EAA00824@smc.vnet.net>

Dear Nathan,
Try out something like:
Nest[Integrate[(# /. {n -> n/x}), {x, a, n/a}, Assumptions -> {a > 0, n 
 > a^2}] &, n - a, 2]

Notice that I extrapolated the assumptions from your mail. If they are 
not correct you might have to change them (but they give the result you 
cited). Also performing this operation for a large muber of iterations 
(in the above code is just 2 but any number will do) can be very slow. 
Maybe you might want to run it once as a NestList and save the output 
for later use.

Jacopo

On 10/11/2011 10:24 AM, Nathan McKenzie wrote:
> I'm working with the following repeated integrals.  Is there any way
> to automate what I'm doing here?
>
> I start with this:
>
> Integrate[ n/x - a, {x, a, n/a}]
>
> The result of that (after a bit of text editing) is a^2 - n Log[a] + n
> (-1 + Log[n/a]). That's the first result I want to work with.  Then, I
> currently manually edit that result by swapping out n with (n/x), and
> have
>
> Integrate[ a^2 - (n/x) Log[a] + (n/x) (-1 + Log[(n/x)/a]), {x, a, n/
> a}]
>
> And that resolves to -a^3 + n Log[a] + n Log[a]^2 + 1/2 n Log[n/a^2]^2
> +  n (a - (1 + Log[a]) Log[n/a]).  Which is the second result I want
> to work with.  The, I manually swap out n with (n/x) and integrate
> again on {x, a, n/a}, and repeat this process ad nauseum.  It doesn't
> take long before the number of n's for me to edit becomes really
> unwieldy and error prone... and ideally I would like to do this many
> times in a row (say, up 30 or 40) and have the results around for
> random use in other contexts.
>
> What I would really like to be able to do is just have some sort of
> function where I can type F[n,a,s], where s is the number of times
> integration is performed, and then n and a (which will be actual
> numbers) will get evaluated.  I feel like the step where I swap out n
> with (n/x) points at something problematic, though.  Is there any way
> in Mathematica for me to construct such a function and get my results
> automatically?
>
>




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