An exact 1-D integration challenge - 66 - (sqrt, sin)
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From: Axel Vogt on 2 Aug 2010 16:37 clicliclic(a)freenet.de wrote: > Vladimir Bondarenko schrieb: >> int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2); >> >> ? > > Numerically, this integral evaluates to: > > 1.7682569591454495299 > > I have transformed it into a sum of three 3F2(1)'s: > > SQRT(pi)/6*(SUM((2*k-2/3)!/(2*k+5/6)!,k,0,inf) > +SUM((2*k)!/(2*k+3/2)!,k,0,inf) > +SUM((2*k+2/3)!/(2*k+13/6)!,k,0,inf)) > > - hopefully correctly. > > Martin. Though 3F2(1) seems not simpler to me than elliptic functions: would you mind to sketch your way?
From: clicliclic on 3 Aug 2010 03:09 Axel Vogt schrieb: > > clicliclic(a)freenet.de wrote: > > > > Vladimir Bondarenko schrieb: > >> > >> int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2); > >> > >> ? > > > > Numerically, this integral evaluates to: > > > > 1.7682569591454495299 > > > > I have transformed it into a sum of three 3F2(1)'s: > > > > SQRT(pi)/6*(SUM((2*k-2/3)!/(2*k+5/6)!,k,0,inf) > > +SUM((2*k)!/(2*k+3/2)!,k,0,inf) > > +SUM((2*k+2/3)!/(2*k+13/6)!,k,0,inf)) > > > > - hopefully correctly. > > > > Martin. > > Though 3F2(1) seems not simpler to me than elliptic > functions: would you mind to sketch your way? I have just followed my proposal to apply Meijer-G convolution, where I have represented SQRT(1-t^3) in INT(SQRT(1-t^3)/(1-t^2),t,0,1) by a G function that vanishes for |t| > 1. Because one has to transform to G(t^6), the order of the functions becomes fairly high. The result in terms of a Meijer-G function reads: -pi^(3/2) / (4*SQRT(2)) * MEIJER([1/6, 1/2, 5/6], [-1/3, 1/3, 3/4, 5/4], [1/6, 1/2, 1/2, 5/6], [-1/3, 0, 1/3], 1) I had hoped that Maple or MMA would see a way to simplify this because of the high symmetry in the three hypergeometric sums: summand numerators are GAMMA(2*k + [1, 3, 5]/3); denominators are GAMMA(2*k + [1, 3, 5]/3 + 3/2). But since Derive doesn't use convergence acceleration to numerically evaluate infinite sums, I couldn't even quickly do a satisfactory sanity check. Martin.
From: Axel Vogt on 3 Aug 2010 09:49 clicliclic(a)freenet.de wrote: > Axel Vogt schrieb: >> clicliclic(a)freenet.de wrote: >>> Vladimir Bondarenko schrieb: >>>> int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2); >>>> >>>> ? >>> Numerically, this integral evaluates to: >>> >>> 1.7682569591454495299 >>> >>> I have transformed it into a sum of three 3F2(1)'s: >>> >>> SQRT(pi)/6*(SUM((2*k-2/3)!/(2*k+5/6)!,k,0,inf) >>> +SUM((2*k)!/(2*k+3/2)!,k,0,inf) >>> +SUM((2*k+2/3)!/(2*k+13/6)!,k,0,inf)) >>> >>> - hopefully correctly. >>> >>> Martin. >> Though 3F2(1) seems not simpler to me than elliptic >> functions: would you mind to sketch your way? > > I have just followed my proposal to apply Meijer-G convolution, where I > have represented SQRT(1-t^3) in INT(SQRT(1-t^3)/(1-t^2),t,0,1) by a G > function that vanishes for |t| > 1. Because one has to transform to > G(t^6), the order of the functions becomes fairly high. The result in > terms of a Meijer-G function reads: > > -pi^(3/2) / (4*SQRT(2)) * MEIJER([1/6, 1/2, 5/6], [-1/3, 1/3, 3/4, 5/4], > [1/6, 1/2, 1/2, 5/6], [-1/3, 0, 1/3], 1) > > I had hoped that Maple or MMA would see a way to simplify this because > of the high symmetry in the three hypergeometric sums: summand > numerators are GAMMA(2*k + [1, 3, 5]/3); denominators are GAMMA(2*k + > [1, 3, 5]/3 + 3/2). But since Derive doesn't use convergence > acceleration to numerically evaluate infinite sums, I couldn't even > quickly do a satisfactory sanity check. > > Martin. Maple confirms it numerically and I think you are correct (though I am not reasonable familiarness with MeijerG): Maple re-writes the term by convert((1-t^3)^(1/2), MeijerG, include=all). Now integrate t^(2*k)*theAbove, giving in H = 1/3*1/(2/3*k+1/3)*t^(2*k+1) *hypergeom([-1/2, 2/3*k+1/3],[4/3+2/3*k],t^3) Evaluate in t=0 equals 0 and for t=1 that can be simplified (just use the Maple command) to 1/6*Pi^(1/2)*GAMMA(2/3*k+1/3)/GAMMA(11/6+2/3*k). Let be u(k) that term, a quotient of two Gammas. Then Sum(u(k), k=0..infinity) numerically gives the desired value (Maple has the needed convergence accelerations). To see your representation better use Sum(u(k)*z^k, k=0..infinity) which Maple can find as a sum of 3 pieces of 3F2 directly. Or looking u(3*k)*t^(3*k) + u(3*k+1)*t^(3*k+1) + u(3*k+2)*t^(3*k+2) and summing each summand shows it as well individually. In the above 1/(1-t^2) = Sum(t^(2*k),k=0..infinity) and interchanging summation and integration was used.
From: Axel Vogt on 3 Aug 2010 09:51 Axel Vogt wrote: > clicliclic(a)freenet.de wrote: >> Axel Vogt schrieb: >>> clicliclic(a)freenet.de wrote: >>>> Vladimir Bondarenko schrieb: Hm ... my Germish usually is better after more coffee.
From: clicliclic on 3 Aug 2010 09:53
clicliclic(a)freenet.de schrieb: > > Vladimir Bondarenko schrieb: > > > > int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2); > > > > ? > > Numerically, this integral evaluates to: > > 1.7682569591454495299 > > I have transformed it into a sum of three 3F2(1)'s: > > SQRT(pi)/6*(SUM((2*k-2/3)!/(2*k+5/6)!,k,0,inf) > +SUM((2*k)!/(2*k+3/2)!,k,0,inf) > +SUM((2*k+2/3)!/(2*k+13/6)!,k,0,inf)) > > - hopefully correctly. > This result can be improved upon as follows. We split the integral as: 1/2*INT(SQRT(1-t^3)/(1-t),t,0,1)+1/2*INT(SQRT(1-t^3)/(1+t),t,0,1) Since the Meijer-G functions need only be transformed to G(t^3) now, their order will be lower. The first term gives three 2F1(1)'s, which can of course be evaluated in closed form: SQRT(pi)*(1/3)!/(4*(5/6)!)*F21(1,1/3,11/6,1)+3*SQRT(pi)*(2/3)!/(~ 28*(1/6)!)*F21(1,2/3,13/6,1)+1/9*F21(1,1,5/2,1) 9*(2/3)!*(5/6)!/(10*SQRT(pi))+3*(1/3)!*(1/6)!/(2*SQRT(pi))+1/3 1.4656093125343588864 The second term can be transformed into three 2F1(-1)'s, one of which is clearly elementary: SQRT(pi)*(1/3)!/(4*(5/6)!)*F21(1,1/3,11/6,-1)-3*SQRT(pi)*(2/3)!/~ (28*(1/6)!)*F21(1,2/3,13/6,-1)+1/9*F21(1,1,5/2,-1) SQRT(pi)/12*SUM((k-2/3)!*COS(pi*k)/(k+5/6)!,k,0,inf)-SQRT(pi)/12~ *SUM((k-1/3)!*COS(pi*k)/(k+7/6)!,k,0,inf)+SQRT(pi)/12*SUM(k!*COS~ (pi*k)/(k+3/2)!,k,0,inf) SQRT(pi)/12*2.484779611412182053-SQRT(pi)/12*0.99195099031185643~ 773+(SQRT(2)*LN(SQRT(2)+1)-1)/3 0.30264764661109064355 The numerical values for the other two sums are derived from integral representations. Martin. |