Re: Multiple integration of UnitStep fails

*To*: mathgroup at smc.vnet.net*Subject*: [mg63321] Re: Multiple integration of UnitStep fails*From*: Maxim <m.r at inbox.ru>*Date*: Fri, 23 Dec 2005 05:08:42 -0500 (EST)*References*: <do0n3j$4ku$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Sat, 17 Dec 2005 09:51:47 +0000 (UTC), Dr. Wolfgang Hintze <weh at snafu.de> wrote: > Hello group, > > trying to solve the nice problem of determining the probability pn that > a polygon formed by n (>=4) random points on the unit circle is void of > an acute angle I came up with the following multiple integral (written > down here for n=5) > > In[15]:= > p5 = (4!*Integrate[Integrate[Integrate[Integrate[UnitStep[\[Phi]4 - Pi, > \[Phi]5 - \[Phi]2 - Pi, Pi - \[Phi]3, Pi - \[Phi]4 + \[Phi]2, > Pi - \[Phi]5 + \[Phi]3], {\[Phi]5, \[Phi]4, 2*Pi}], {\[Phi]4, > \[Phi]3, 2*Pi}], {\[Phi]3, \[Phi]2, 2*Pi}], {\[Phi]2, 0, 2*Pi}])/(2*Pi)^4 > > Mathematica version 4 was not able to solve this but returned it > unevaluated after some minutes; version 5 complained several things > like: argument is not a power series, unable to check convergence, but > didn't come up with any result in ten minutes (I wouldn't wait longer). > > I could successfully check the normalization at least: > > In[14]:= > p5 = (4!*Integrate[Integrate[Integrate[Integrate[UnitStep[1], {\[Phi]5, > \[Phi]4, 2*Pi}], {\[Phi]4, \[Phi]3, 2*Pi}], {\[Phi]3, \[Phi]2, 2*Pi}], > {\[Phi]2, 0, 2*Pi}])/(2*Pi)^4 > > Out[14]= 1 > > How would you proceed to solve In[15]? What about the general case (n=6 > see below)? > > Any hints are greatly appreciated. > > Regards, > Wolfgang > > PS: > > Here's the probability for the case n=6 > > p6 = > (5!*Integrate[Integrate[Integrate[Integrate[Integrate[UnitStep[\[Phi]5 - > Pi, \[Phi]6 - \[Phi]2 - Pi, Pi - \[Phi]3, Pi - \[Phi]4 + \[Phi]2, > Pi - \[Phi]5 + \[Phi]3, Pi - \[Phi]6 + \[Phi]4], {\[Phi]6, > \[Phi]5, 2*Pi}], {\[Phi]5, \[Phi]4, 2*Pi}], {\[Phi]4, \[Phi]3, 2*Pi}], > {\[Phi]3, \[Phi]2, 2*Pi}], {\[Phi]2, 0, 2*Pi}])/(2*Pi)^5 > > $Aborted > The integral in p6 can be evaluated if we first perform the cylindrical decomposition of the region where UnitStep is non-zero: In[1]:= p6 = Hold[ (5!*Integrate[Integrate[Integrate[Integrate[Integrate[ UnitStep[\[Phi]5 - Pi, \[Phi]6 - \[Phi]2 - Pi, Pi - \[Phi]3, Pi - \[Phi]4 + \[Phi]2, Pi - \[Phi]5 + \[Phi]3, Pi - \[Phi]6 + \[Phi]4], {\[Phi]6, \[Phi]5, 2*Pi}], {\[Phi]5, \[Phi]4, 2*Pi}], {\[Phi]4, \[Phi]3, 2*Pi}], {\[Phi]3, \[Phi]2, 2*Pi}], {\[Phi]2, 0, 2*Pi}])/(2*Pi)^5]; p6 //. HoldPattern[Integrate[Integrate[e_, Siter2__], Siter__]] :> Integrate[e, Siter, Siter2] /. HoldPattern[Integrate[UnitStep[s__], Siter__]] :> Integrate[Boole[Reduce[ Join[(# >= 0&) /@ {s}, (#[[2]] <= #[[1]] <= #[[3]]&) /@ {Siter}], {Siter}[[All, 1]], Reals]], Sequence @@ ({#[[1]], -Infinity, Infinity}&) /@ {Siter}] // ReleaseHold Out[2]= 1/4 With PiecewiseIntegrate ( http://library.wolfram.com/infocenter/MathSource/5117/ ) we only need to carry out the first step: In[3]:= p6 //. HoldPattern[Integrate[Integrate[e_, Siter2__], Siter__]] :> Integrate[e, Siter, Siter2] /. Integrate -> PiecewiseIntegrate // ReleaseHold Out[3]= 1/4 Another thing that often works for piecewise integrands is to restrict the domain of the parameter values: In[4]:= (int = Integrate[Min[x, y], {x, 0, a}, {y, 0, b}];) // Timing Out[4]= {114.547*Second, Null} In[5]:= Cases[int, If[a_, __] :> a, {0, -1}] Out[5]= {b > 0 && (b <= Re[a] || Re[a] <= 0 || Im[a] != 0)} In[6]:= int /. {{a -> 2, b -> 1}, {a -> 2, b -> 3}} Out[6]= {Indeterminate, 6} We can see that the condition in the answer allows for the possibility of complex values (for a but not b), which I don't think is in any way useful, and even at that the conditional answer is incorrect for {a, b} = {2, 1}. Besides, the condition doesn't cover all the possibilities, so for {a, b} = {2, 3} the integration has to be performed again, and the result is also incorrect. With the assumption Element[{a, b}, Reals] the integration takes significantly less time and we get an answer that is correct for all values of the parameters. Maxim Rytin m.r at inbox.ru