R: R: Re: Difficult antiderivative

*To*: mathgroup at smc.vnet.net*Subject*: [mg128949] R: R: Re: Difficult antiderivative*From*: "Brambilla Roberto Luigi (RSE)" <Roberto.Brambilla at rse-web.it>*Date*: Wed, 5 Dec 2012 03:13:38 -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> <20121130105825.AB3E468E2@smc.vnet.net> <0C6D72B120D0634E927CAA8661F84B75535FE7@welkrock.ricerca.lan> <50BDCCA1.4070801@uos.de>

Dear Roland, given the integral (1) Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x] (0<r<a) Applying your substitution x->r Cos[u] I have found (p=a/r<1) (2) Integrate[ ArcCosh[ p Sec[u] ],u] With your previous substitution x-> a/Cosh[u] I have found (3) -p(NIntegrate[u Tanh[u]/Sqrt[Cosh[u]^2 -p^2],u] And verified that (1),(2) and (3) give the same results when numerically eva luated on a finite interval (b,r) with (b<r<a). Unfortunately none of them seems to be solvable in the realm (rather large!) of the Mathematica8 funct ions. May be one has to use higher order trascendent functions? Or, wisely, stop h ere? Many thanks again for your attention, Rob -----Messaggio originale----- Da: Franzius, R. [mailto:roland.franzius at uos.de] Inviato: marted=EC 4 dicembre 2012 11.13 A: Brambilla Roberto Luigi (RSE) Oggetto: Re: R: Re: Difficult antiderivative Am 30.11.2012 14:45, schrieb Brambilla Roberto Luigi (RSE): > Dear Roland, > > Many thanks for your suggestion, i.e. given the indefinite integral (antid erivative) with r<a : > > 1) Integrate[ ArcCosh[a/x]/Sqrt[r^2-x^2],x] > > (that Mathematica8 can't solve) change the variable x->a/Cosh[u]. > Doing the substitution the integral becomes > > 2) Integrate[ u Tanh[u]/Sqrt[q^2 Cosh[u]^2-1],x] > > where q=r/a (r<a). Unfortunately also this integral is unsolvable by > Mathematica8 (unless q=1). > > Alexei Boulbitch wrote (29 november) [mg128833] > .... > most of indefinite integrals have this property ("does not exist") , and o nly smaller part of them can be expressed in terms of analytical and special functions. > .... > For some integrals you can find the solution, for others you cannot, and n o general rule exists that would help you to distinguish one group from the other. > > Are (1) and (2) cases of this unhappy class? > There was probably a typo somewhere. I didn't store the procedure and cam't reproduce the complete polylog results. The most simple algebraic form may be replace Sqrt(r^2-x^2) /. x->r Cos[u] to get a Sin[u] which is cancled by the derivative So you are left with Integrate[ ArcCosh[ p Csc[u] ], u ] This function is a bit strange, the Argument of ArcCosh has to be >1 demanding 0 < Cos[u] < Min[1,p] Apply TrigToExp and with some manipulations on the Log you get a term Integrate[Log[Cos[u]],u ] an insolvable rest which my look like Integrate[ Log[1 + Sqrt[ 1- q^2 Cos[u]^2] ], u ] Regards Roland Franzius RSE SpA ha adottato il Modello Organizzativo ai sensi del D.Lgs.231/2001, in forza del quale l'assunzione di obbligazioni da parte della Societ=E0 avvie ne con firma di un procuratore, munito di idonei poteri. RSE adopts a Compliance Programme under the Italian Law (D.Lgs.231/2001). Ac cording to this RSE Compliance Programme, any commitment of RSE is taken by the signature of one Representative granted by a proper Power of Attorney. Le informazioni contenute in questo messaggio di posta elettronica sono ris ervate e confidenziali e ne e' vietata la diffusione in qualsiasi modo o for ma. Qualora Lei non fosse la persona destinataria del presente messaggio, La invitiamo a non diffonderlo e ad eliminarlo, dandone gentilmente comunicazi one al mittente. The information included in this e-mail and any attachments are confidential and may also be privileged. If you are not the correct rec ipient, you are kindly requested to notify the sender immediately, to cancel it and not to disclose the contents to any other person.