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Re: six-fold integral


Your result 

(2 3^(1/2)-2^(1/2)-1)/5 + pi/3 + ln[(2^(1/2)-1)(2-3^(1/2)]

misses a closing ')'. Assuming you mean 

(2 3^(1/2)-2^(1/2)-1)/5+Pi/3 + Log[(2^(1/2)-1)(2-3^(1/2))] // N,

this is unfortunately negative (-0.941156).

Using the joint probability density for (x - y) it is straightforward
to calculate the following exact result in just a few lines:

In[1]:= logRewrite[expr_] := FixedPoint[PowerExpand[# //.
         Log[a_] :> Log[Factor[Together[a]]]]&, TrigToExp[expr]]

In[2]:= indefiniteIntegral = Integrate[# 2(1 - z), z]& /@
Expand[logRewrite[
   Integrate[Integrate[1/Sqrt[x^2 + y^2 + z^2] 2(1 - x), x] 2(1 - y),
y]]];

In[3]:= replaceLimits[expr_, xyz_] :=((# /. xyz -> 1) - (# /. xyz ->
0))&[
 Expand[logRewrite[expr]] /. xyz^n_?Positive Log[xyz] -> 0]

In[4]:= Fold[replaceLimits, indefiniteIntegral, {z, y, x}] //
Together;

In[5]:= FullSimplify[%]

Out[5]= -((2*Pi)/3) + Log[3 + 2*Sqrt[2]] + 
  1/10*(4 + 4*Sqrt[2] - 8*Sqrt[3] - 25*Log[2] + 50*Log[1 + Sqrt[3]] - 
    5*Log[2 + Sqrt[3]])

In[6]:= N[%]

Out[6]= 1.88231

This agrees with a Monte-Carlo check:

In[7]:=
SeedRandom[888];
Compile[{{n, _Integer}},
        Module[{sum = 0.},
             Do[sum = sum + 1/Sqrt[(Random[] - Random[])^2 +
                                   (Random[] - Random[])^2 +
                                   (Random[] - Random[])^2],
                {n}]; sum/n]][10^6]

Out[7]= 1.88295


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