Re: Manually tell Mathematica how to evaluate integrals

*To*: mathgroup at smc.vnet.net*Subject*: [mg128804] Re: Manually tell Mathematica how to evaluate integrals*From*: Hui <e.schlemm at hotmail.de>*Date*: Tue, 27 Nov 2012 03:33:09 -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*: <k8sqlr$ooc$1@smc.vnet.net> <k8uro7$st9$1@smc.vnet.net>

Thank you DC. There is a typo in my original statement. I meant to suggest that x PolyLog[n+1,Exp[x]] - PolyLog[n+2,Exp[x] is the anti-derivative of the function x PolyLog[n,Exp[x]]. That seems to be confirmed by differentiating the former expression. Any ideas as to why Mathematica won't evaluate this integral, even in the explicit case of, say, n=4? On Monday, November 26, 2012 4:40:54 AM UTC, DC wrote: > The following doesn't seem to reproduce your statement : > > > > Simplify[D[x PolyLog[n + 1, Exp[x]] - x PolyLog[n + 2, Exp[x]], x], > > Assumptions -> {n \[Element] Integers, x \[Element] Reals}] > > > > > > > > On Sunday, 25 November 2012 10:10:17 UTC, Hui wrote: > > > Hi all, > > > > > > > > > > > > I have a question about Mathematica's abilities to solve integrals. There seem to be cases where an antiderivative is explicitly known yet Mathematica fails to compute the integral. One example of this would be > > > > > > > > > > > > Integrate[x PolyLog[n,Exp[x]],x] > > > > > > > > > > > > which Mathematica only solves for n=1,2, even though it is quite easy to find a solution for higher values of n as well, namely > > > > > > > > > > > > x PolyLog[n+1,Exp[x]] - x PolyLog[n+2,Exp[x]. > > > > > > > > > > > > I would like to know if it possible to teach Mathematica this integral in such a way that it will also recognise and solve it in more complicated cases such as > > > > > > > > > > > > Integrate[(x+a) PolyLog[n,b Exp[c x]],x]. > > > > > > > > > > > > Thank you very much, your assistance is much appreciated! > > > > > > > > > > > > Hui