Re: Difficult antiderivative

*To*: mathgroup at smc.vnet.net*Subject*: [mg128853] Re: Difficult antiderivative*From*: Roland Franzius <roland.franzius at uos.de>*Date*: Fri, 30 Nov 2012 05:58:25 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <k94hra$bk4$1@smc.vnet.net>

Am 28.11.2012 09:26, schrieb Brambilla Roberto Luigi (RSE): > Using Mathematica (v5.1 to v.8) it is possible to obtain (in few > seconds) the following antiderivative > > > > (1) Integrate[ x ArcCosh[a/x]/Sqrt[r^2-x^2],x] > > > > (a rather lengthy expression). > > > > It happens that I need > > > > (2) Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x] > or equivalently > > > > (3) Integrate[ x^2 ArcCosh[a/x]/Sqrt[r^2-x^2],x] > I have tried a lot of tricks (by part integration etc..) but I was not > able to find it. > > Q.: may be that the antiderivative does not exist? > > The numerical integral (b<r<a) > > > > NIntegrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],{x,b,r}] > > Don't give any problem. x/Sqrt[a-x^2 ] is a simple derivative of Sqrt[a-x^2]. Substitutuing x-> a/Cosh[u], with the right assumptions about r>x>0, I get an expression with Logs and Polylogs. But itd to tedious to check it by resubstitution and FullSimplifying the derivative. A crucial point is to replace Sqrt[r^2- a^2/Cosh[u]^2 ] -> Sqrt[q^2 Cosh[u]^2-1] in order to get a managable expression. -- Roland Franzius