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MathGroup Archive 2000

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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


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