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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: [mg22004] Re: Why doesn't Mathematica know this?
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Thu, 10 Feb 2000 02:25:46 -0500 (EST)
  • Organization: Wolfram Research, Inc.
  • References: <87lu84$6q9@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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?


For the numerical check it is typically better to use NIntegrate and to
give PrecisionGoal/WorkingPrecision option values to control the quality
of the result. For example, by upping the PrecisionGoal (default is 6
for dfault NIntegrate method) I can get fairly close to the numerical
result shown above for the PolyLog.

In[17]:= Timing[4/Sqrt[Pi]*NIntegrate[x^2*Log[1 + 1*Exp[-x^2]],
{x,0,Infinity}]] // InputForm
Out[17]//InputForm= {0.019999999999999574*Second, 0.8671998802871057}

In[19]:= Timing[4/Sqrt[Pi]*NIntegrate[x^2*Log[1 + 1*Exp[-x^2]],
{x,0,Infinity}, PrecisionGoal->12]] // InputForm
Out[19]//InputForm= {0.07000000000000028*Second, 0.8671998890121846}

Without much trouble I can get essentially full agreement.

In[24]:= Timing[4/Sqrt[Pi]*NIntegrate[x^2*Log[1 + 1*Exp[-x^2]],
{x,0,Infinity}, PrecisionGoal->20, WorkingPrecision->30]] // InputForm
Out[24]//InputForm= 
{1.3199999999999994*Second,
0.86719988901218413819134718586552489872`20.6973}

In[25]:= Timing[N[-PolyLog[5/2, -1], 20]] // InputForm
Out[25]//InputForm= {0.07000000000000028*Second,
0.867199889012184138191347179660941`20}

Note that these agree well into the guard digits (that is, past the
stated precision of 20 digits).


As to whether or why the integral should fail, I cannot help.


To show the desired identity here are two useful methods.

(i) Show the functions agree at one point and have equal derivatives.

In[12]:= PolyLog[5/2, 0]
Out[12]= 0

In[13]:= 4/Sqrt[Pi]*Integrate[x^2*Log[1], {x, 0, Infinity}]
Out[13]= 0

In[15]:= InputForm[D[-PolyLog[5/2, -z], z]]
Out[15]//InputForm= -(PolyLog[3/2, -z]/z)

In[16]:= InputForm[Simplify[D[4/Sqrt[Pi]*Integrate[x^2*Log[1 +
z*Exp[-(x^2)]],{x, 0, Infinity}], z]]]
Out[16]//InputForm= -(PolyLog[3/2, -z]/z)


(ii) Show the series expansions are equal.

In[2]:= <<DiscreteMath`RSolve`

In[7]:= SeriesTerm[PolyLog[5/2, -z], {z, 0, n}] // InputForm
Out[7]//InputForm= ((-1)^n*UnitStep[-1 + n])/n^(5/2)

In[8]:= Integrate[SeriesTerm[4/Sqrt[Pi]*x^2*Log[1 + z*Exp[-(x^2)]],
{z,0,n}], {x,0,Infinity}, Assumptions->n>0] // InputForm
Out[8]//InputForm= -(((-1)^n*UnitStep[-1 + n])/n^(5/2))


Daniel Lichtblau
Wolfram Research


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