
Re: integral for fun
Posted:
Mar 3, 2014 2:30 AM


On Sunday, March 2, 2014 12:29:38 PM UTC+2, clicl...@freenet.de wrote: > Jonas Matuzas schrieb: > > > On Saturday, March 1, 2014 1:51:38 AM UTC+2, Nasser M. Abbasi wrote: > > > > On 2/28/2014 3:39 PM, Axel Vogt wrote: > > > > > > > > > > Maple 17 actually gives it 'immediately' in terms of hypergeometric 0F7 > > > > > (and your numerical value) > > > > > > > > > > > > > Mathematica actually finds the indefinite integral with no > > > > problem. The problem comes when evaluating this result at > > > > the limits of integration in order to find the final value. > > > > > > > > The limit has to be used, becuase directly evaluation of x=Infinity > > > > or x=Infinity producdes indeterminate form(s). > > > > > > > > When trying the limit, Mathematica is taking for ever to finish, > > > > so I stopped it after many hrs. (same what happens it seems > > > > with the OP) > > > > > > > > int = Integrate[Cos[x]^4/(1 + x^8), x]; > > > > > > > > int/.x>Infinity > > > > > > > > Infinity::indet: Indeterminate expression 1+\[Infinity]+\[Infinity] encountered. >> > > > > > > > > While > > > > > > > > Limit[int,x>Infinity] (* takes long time....*) > > > > > > > > And I could not figure a way around it in the small amount of time I played > > > > with it. Here is "int" fyi: It contains many expressions involving > > > > the CosIntegral and SinIntegral (It seems Maple used hypergeometric 0F7 > > > > > > > > https://reference.wolfram.com/mathematica/ref/SinIntegral.html > > > > http://reference.wolfram.com/mathematica/ref/CosIntegral.html > > > > > > > >  > > > > > > > > (3/32)*ArcTan[Sec[Pi/8]*(x  Sin[Pi/8])]* > > > > Cos[Pi/8] + (3/32)* > > > > ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])]* > > > > Cos[Pi/8]  (3/64)*Cos[Pi/8]* > > > > Log[1 + x^2  2*x*Cos[Pi/8]] + > > > > (3/64)*Cos[Pi/8]*Log[1 + x^2 + > > > > 2*x*Cos[Pi/8]] + > > > > (3/32)*ArcTan[(x  Cos[Pi/8])*Csc[Pi/8]]* > > > > Sin[Pi/8] + (3/32)*ArcTan[(x + Cos[Pi/8])* > > > > Csc[Pi/8]]*Sin[Pi/8]  > > > > (3/64)*Log[1 + x^2  2*x*Sin[Pi/8]]* > > > > Sin[Pi/8] + (3/64)* > > > > Log[1 + x^2 + 2*x*Sin[Pi/8]]*Sin[Pi/8] + > > > > (1/2)*((((1)^(7/8)*(Cos[2*(1)^(1/8)]* > > > > CosIntegral[2*(1)^(1/8) + 2*x] + > > > > Sin[2*(1)^(1/8)]*SinIntegral[ > > > > > > > > .... too large to post it all > > > > > > > > Maple 17 also used Si and Ci, but it uses summation in > > > > its output, hence it looks much smaller than the output above > > > > > > > >  > > > > > > > > z:=int(cos(x)^4/(1 + x^8), x); > > > > > > > > (1/32)*(sum((8192*(Si(4*x+_R1)*sin(_R1)+Ci(4*x_R1)* > > > > cos(_R1)))/_R1^7, _R1 = RootOf(_Z^8+65536)))+(1/4)* > > > > (sum((32*(Si(2*x+_R1)*sin(_R1)+Ci(2*x_R1)* > > > > cos(_R1)))/_R1^7, _R1 = RootOf(_Z^8+256)))+(3/8)* > > > > (sum(_R*ln(x+8*_R), _R = RootOf(16777216*_Z^8+1))) > > > > > > > >  > > > > > > > > > > thank you for your analysis . You think Matematics is calculating limits? I think it should use Residues theory... > > > > Typically, Mathematica and Maple would here split the integration > > interval and use convolution of MeijerG functions for INT(COS(c*x)/(1 + > > x^8), x, 0, inf), I think. As Axel has noted, COS(x)^8 = 1/8*cos(4*x) + > > 1/2*cos(2*x) + 3/8 can be decomposed accordingly. The MeijerG technique > > allows to deal a large range of elementary and nonelementary factors in > > the integrand and can be implemented as a sequence of transformation and > > simplification rules. The 0F7 expressions were presumably obtained in > > this way, although their ultimate reducibility to elementary > > constituents may be difficult to detect, and the reduction difficult to > > effect. See <http://en.wikipedia.org/wiki/Meijer_Gfunction> for some > > background. > > > > Martin.
yes.. I know these functions. Still calculating!!! it is already 4'th day has started (start was Friday morning). I am using 6'th fastest cpu in the world http://www.cpubenchmark.net/singleThread.html Is is worth to wait for the end ? My question is what is he (mathematica) doing ? Is it fall in eternal loop ?

