Re: six-fold integral
- To: mathgroup at smc.vnet.net
- Subject: [mg30098] Re: six-fold integral
- From: mtrott at wolfram.com (Michael Trott)
- Date: Sat, 28 Jul 2001 01:51:00 -0400 (EDT)
- References: <9jr6uk$4mn$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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