Re: Making a function out of repeated hyperbola integrations?

*To*: mathgroup at smc.vnet.net*Subject*: [mg122077] Re: Making a function out of repeated hyperbola integrations?*From*: Patrick Scheibe <pscheibe at trm.uni-leipzig.de>*Date*: Wed, 12 Oct 2011 03:43:45 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201110110824.EAA00824@smc.vnet.net>

Hi, 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: NestList[Function[ 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 list. Cheers Patrick 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? >

**References**:**Making a function out of repeated hyperbola integrations?***From:*Nathan McKenzie <kenzidelx@gmail.com>