Re: Why doesn't Mathematica know this?
- To: mathgroup at smc.vnet.net
- Subject: [mg21976] Re: Why doesn't Mathematica know this?
- From: Roland Franzius <Roland.Franzius at uos.de>
- Date: Mon, 7 Feb 2000 13:02:29 -0500 (EST)
- References: <87lu84$6q9@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hello Ginsand
In: q = Series[Log[1 + x Exp[-y]], {x, 0, 5}]
Out: SeriesData[x, 0, {E^(-y), -1/(2*E^(2*y)),
1/(3*E^(3*y)), -1/(4*E^(4*y)), 1/(5*E^(5*y))}, 1, 6, 1]
In: h[x_,y_]=Normal[q]
(1/5*x^5)/E^(5*y) - (1/4*x^4)/E^(4*y) + (1/3*x^3)/E^(3*y) -
(1/2*x^2)/E^(2*y) + x/E^y
In: -2/Sqrt[Pi]Integrate[Sqrt[y]h[x, y], {y, 0, Infinity}] //
Expand
Out: -x + x^2/(4*Sqrt[2]) - x^3/(9*Sqrt[3]) + x^4/32 -
x^5/(25*Sqrt[5])
In: Series[PolyLog[5/2, -x], {x, 0, 5}]
Out: -x + x^2/(4*Sqrt[2]) - x^3/(9*Sqrt[3]) + x^4/32 -
x^5/(25*Sqrt[5])
Try a proof along this line.
Looking up Prudnikov/Brychkov/Marichev, Integrals and Series
we find a single formula for Integrals of the form
Integrate[ x^a Log(b+ce^(d x))]
2.6.28 Integate[x^n Log[1-Exp[-a x]],{x,0,Infinity}]
= {-(2/a)^(n+1)Pi^(n+2)Abs[B_{n+2}]/((n+2)(n+1)) /; a>0, n
Even
{-n!a^(-n-1) RiemannZeta[n+2] /; a>0, n Odd
This is answering the question why Mathematica doesn't know an
answer. Where?
regards roland
ginsand 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?
--
Roland Franzius
Theor. Physik FB Physik, Univ. Osnabrueck
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