Re: Why doesn't Mathematica know this?

*To*: mathgroup at smc.vnet.net*Subject*: [mg21977] Re: Why doesn't Mathematica know this?*From*: Matthias Weber <weber at math.uni-bonn.de>*Date*: Mon, 7 Feb 2000 13:02:29 -0500 (EST)*Organization*: RHRZ - University of Bonn (Germany)*References*: <87lu84$6q9@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <87lu84$6q9 at smc.vnet.net>, "ginsand" <gar at none.home> wrote: > Hello all, > > I think this is true: > > 4/Sqrt[Pi]*Integrate[x^2*Log[1 + z*Exp[-(x^2)]],{x, 0, Infinity}] > > Is actually: > > -PolyLog[5/2, -z] > > Yet Mathematica doesn't know that. > Moreover, even if I type a numerical value instead of z in the Integral, > and > evaluate the expression numericaly (//N), Mathematica doesn't generally > converges (a $RecursionLimit message, supressed and hang), except for > special values of z. > > And finally, look at this: > > In[1]:=Timing[N[4/Sqrt[Pi]* > Integrate[x^2*Log[1 + 1*Exp[-x^2]], {x, 0, Infinity}]]] > > Out[1]={22.67999999999999*Second, 0.8671998802871057} > > In[2]:=Timing[N[-PolyLog[5/2, -1]]] > > Out[2]={0.3300000000000054*Second, 0.8671998890121841} > > It isn't exactly the same result... Still I think the equallity should > hold > (It has a known physical meaning). > > > Comments, suggestions or insight on this issue? > > Knowledge about special functions is poor. You can convince yourself of your claim by developing both expressions into a power series at z=0. Mathematica can help finding the (simple) general coefficient: Normal[4/Sqrt[Pi]*Series[x^2*Log[1 + z*Exp[-(x^2)]], {z, 0, 6}]] Integrate[%, {x, 0, Infinity}] gives the same as Series[-PolyLog[5/2, -z], {z, 0, 6}] An actual proof of your identity is easily derived from this. Hope that helps. Matthias