
Re: Manually tell Mathematica how to evaluate integrals
Posted:
Dec 3, 2012 3:18 AM


repl = Integrate[Log[1 + c_*Exp[x]], x] > (Integrate[Log[1 + temp*Exp[x]], x] /. temp > c);
expr1 = Log[1 + Exp[x + a] + Exp[x + b]];
Factor out the Exp[x]
expr2 = Map[Collect[#, Exp[x]] &, expr];
Verifying that the expressios are equivalent
expr1 == expr2 // Simplify
True
Then do exactly as before
expr3 = Integrate[expr2, x] /. repl
PolyLog[2, (E^x)*(E^a + E^b)]
Verifying the result
D[expr3, x] == expr1 // Simplify
True
Bob Hanlon
On Wed, Nov 28, 2012 at 11:33 AM, Eckhard Schlemm <e.schlemm@hotmail.de> wrote: > Thank you Mr Hanlon. Unfortunately your change of variables does not apply > to integrals like > > Integrate[Log[1 + d Exp[x] + Exp[x],x] > > Would there be a way of forcing the integrand to be transformed into the > form > > > Log[1 + d Exp[x]] > > if that is possible? I am particularly thinking of expressions like > > Log[1 + Exp[x+a] + Exp[x+b]] > > for which d = Exp[a] + Exp[b]. > > All these integrals have the same structure, so I think Mathematica should > be able to do them without further tricks. > > Thanks again > > > On 28/11/12 15:56, Bob Hanlon wrote: >> >> It's a nuisance but just do a change of variables >> >> Integrate[Log[1 + (d + 1) Exp[x]] /. >> d > d  1, x] /. d > d + 1 >> >> PolyLog[2, (1 + d) E^x] >> >> Integrate[Log[1 + (d + 1) Exp[x]], x] /. >> >> Integrate[Log[1 + c_* Exp[x]], x] > >> (Integrate[Log[1 + temp* Exp[x]], x] /. >> temp > c) >> >> PolyLog[2, (1 + d) E^x] >> >> >> Bob Hanlon >> >> >> On Wed, Nov 28, 2012 at 3:17 AM, Hui<e.schlemm@hotmail.de> wrote: >>> >>> Similar to my first question, I realised that Mathematica can evaluate >>> the integral >>> >>> Integrate[Log[1 + d Exp[x]],x] >>> >>> but fails to find the antiderivative of the function >>> >>> Log[1 + (d+1) Exp[x]]. >>> >>> I find this quite annoying; does anyone a way around the issue? >>> >>> Any input is much appreciated. >>> Thanks, Hui. >>> >>> Am Dienstag, 27. November 2012 08:38:48 UTC schrieb Hui: >>>> >>>> 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 antiderivative 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 >>> >>> >>> >> >> >

