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Re: Making a function out of repeated hyperbola integrations?

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  • Subject: [mg122077] Re: Making a function out of repeated hyperbola integrations?
  • From: Patrick Scheibe <pscheibe at>
  • Date: Wed, 12 Oct 2011 03:43:45 -0400 (EDT)
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with the starting value n-a everything you have to do is to
iterate the the following two steps again and again

1. replacing n->n/x
2. integrating

and that is as simple as it sounds:

  Integrate[# /. n -> n/x, {x, a, n/a}, 
   GenerateConditions -> False]],
 n - a, 4]

If you want the first starting element dropped, use Rest on the resulting


On Oct 11, 2011, at 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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