15-355: Modern Computer Algebra V. Adamchik Symbolic Integration Victor Adamchik Carnegie Mellon University The Rioboo algorithm Consider the following integral x2 +2x+4 dx x4 -7x 2 +2x+ 17 Since the denominator is square free SquareFreeQ17+2x-7x 2 +x 4, x True we proceed with Trager’s algorithm p =4+2x+x 2; q= 17+2x-7x 2 +x 4; Resultant[q, p-zD[q, x], x] // Factor 21691+4z 22 roots= Union[z/. Solve[%⩵ 0, z]]; args= PolynomialGCD[q, p-#D[q, x]]&/@ roots; res= Inner[#1 Log[#2] &, roots, args, Plus] 1 1 - ⅈ Log(-4-ⅈ)-ⅈx+x 2+ ⅈ Log(-4+ⅈ) +ⅈx+x 2 2 2 However, the integral can be computed without the complex numbers extension V. Adamchik 2 4+2x+x 2 Integrate , x 17+2x-7x 2 +x 4 1 -1-x 1 1+x ArcTan - ArcTan 2 -4+x 2 2 -4+x 2 ■ Idea 1 - Log to Arctan Given a field K of characteristic 0 such that -1∉K, and A,B∈K[x] with B ≠ 0, return a sum of arctangents defined by d A d i A+iB arctan = log d x B d x 2 A-iB We will be using =D to denote that computations are correct under differentiation (or up to a constant). Indeed, A i A+iB arctan =D log B 2 A-iB We use differentiation to eliminate the constant of integration A I A+IB A = 1.234; B= 0.432; ArcTan - Log B 2 A-IB 1.5708+ 0.ⅈ A I A+IB A = -1.234; B= 0.432; ArcTan - Log B 2 A-IB -1.5708+ 0.ⅈ The identity readily follows from the definition of the arctangent function i i i 1-ix arctan(x) = log(1-ix)- log(1+ix) =D log 2 2 2 1+ix Indeed, A B A i 1-i i B-iA i -A i -i B-A i A+iB arctan =D log B = log = log i = log =D log B 2 1+i A 2 B+iA 2 B +A 2 -i B+A 2 A-iB B i Using this idea we can rewrite the result of integration as follows 15-355: Modern Computer Algebra V. Adamchik i i i x2 +i(x+1)-4 x2 -4 logx2 +ix + (-4+i)- logx2 -ix + (-4-i)= D log =D arctan 2 2 2 x2 -i(x+1)-4 x+1 Therefore, x2 +2x+4 x2 -4 dx= arctan +C x4 -7x 2 +2x+ 17 x+1 x2 -4 DArcTan , x// Simplify x+1 4+2x+x 2 17+2x-7x 2 +x 4 which matches Mathematica’s result up to the constant of integration 4+2x+x 2 Integrate , x// Simplify 17+2x-7x 2 +x 4 1+x ArcTan 4-x 2 Note, 1 DArcTan[x] + ArcTan , x// Together x 0 It follows, that both results are correct under differentiation x2 +2x+4 x2 -4 x+1 dx= arctan +C 1 = arctan +C 2 x4 -7x 2 +2x+ 17 x+1 4-x 2 ■ Definite Integration The proposed simplification causes computational problems for a definite integration, since an arctan- gent of a rational function contains hidden singularities V. Adamchik 4 x+1 PlotArcTan ,{x, 1, 3} 4-x 2 1.5 1.0 0.5 1.5 2.0 2.5 3.0 -0.5 -1.0 -1.5 We all know that proper definite integrals are evaluated by means of the Newton-Leibniz theorem b f(x)dx=F(b)-F(a) a where F(x) is an antiderivative. In case of the integral discussed above, we get 4 x2 +2x+4 x+1 x+1 dx= arctan x=4 - arctan x=0 4 2 2 2 0 x -7x +2x+ 17 4-x 4-x And this is simply wrong x+1 res= ArcTan ; 4-x 2 (res/. x→4)-(res/. x→0) //N -0.63977 x2 +2x+4 NIntegrate ,{x, 0, 4} x4 -7x 2 +2x+ 17 2.50182 The problem is that the path of integration (0, 4) crosses singularities of the antiderivative. The follow- ing picture demonstrates this 15-355: Modern Computer Algebra V. Adamchik x+1 ContourPlotEvaluateReArcTan /. x ->x+Iy, 4-x 2 {x,-3, 4},{y,-3, 4}, ContourShading -> False, Contours -> 20, PlotPoints -> 40, Epilog -> {Hue[0], Thickness[0.005], Line[{{0, 0},{4, 0}}], RGBColor[0, 0, 0], Thickness[0.003], Line[{{1.95,-0.756},{1.95, 0.756}}], Line[{{-1.95,-0.246},{-1.95, 0.246}}] } 4 3 2 1 0 -1 -2 -3 -3 -2 -1 0 1 2 3 4 Observe that Mathematica computes the definite integral correctly x2 +2x+4 Integrate ,{x, 0, 4}//N x4 -7x 2 +2x+ 17 2.50182 How does Mathematica do it? By computing limits at singularities V. Adamchik 6 Limit[res, x→ 4, Direction ->1]- Limit[res, x→ 2, Direction -> -1] + Limit[res, x→ 2, Direction ->1]- Limit[res, x→ 0, Direction -> -1] //N 2.50182 There is a better idea! We can convert logs into arctangents of polynomials. ■ Idea 2 The “identity” (under differentiation) we used in the previous section A i A+iB arctan =D log B 2 A-iB can be further expanded A i A+iB A d+Bc i d+ic arctan =D log =D arctan + log B 2 A-iB g 2 d-ic where g=Bd-Ac, deg(d) < deg(A), deg(c) < deg(A) Proof. We will simplify log d+i c by using d-i c A i A+iB arctan =D log B 2 A-iB We get i d+ic d log =D arctan 2 d-ic c Thus, A d+Bc i d+ic A d+Bc d arctan + log =D arctan + arctan B d-Ac 2 d-ic B d-Ac c Next, we apply the summation formula for arctangents x+y arctan(x) + arctan(y) = arctan 1-xy 15-355: Modern Computer Algebra V. Adamchik Clear[A, B, c, d]; A d+Bc d x+y x= ; y= ; Simplify B d-Ac c 1-xy B - A We obtain, A d+Bc d B A arctan + arctan =D arctan - =D arctan g c A B whih concludes the proof. QED ■ Example Consider the result of Trager’s algorihm x2 +2x+4 i x2 -4+i(x+1) dx= log +C x4 -7x 2 +2x+ 17 2 x2 -4-i(x+1) We will convert the complex log function into a sum of arctangents by recursively applying this i A+iB A d+Bc i d+ic log =D arctan + log 2 A-iB g 2 d-ic First we find g by running the EEA Clear[x]; A =x 2 - 4; B=x+ 1; {g,{d, c}} = PolynomialExtendedGCD[B,-A, x] 1 1 1, (-1+x), 3 3 and then compute V. Adamchik 8 A d+Bc I d+Ic ArcTan // Simplify+ Log g 2 d-Ic ⅈ + 1 (- + ) 1 1 1 x ArcTan 5-3x-x 2 +x 3+ ⅈ Log 3 3 ⅈ 3 2 - + 1 (-1+x) 3 3 to get the following x-1 i i x2 -4+i(x+1) 1 i + log =D arctan x3 -x 2 -3x+5 + log 3 3 2 x2 -4-i(x+1) 3 2 x-1 - i 3 3 Apply the same procedure to the remaining log x-1 1 A = ;B= ; 3 3 {g,{d, c}} = PolynomialExtendedGCD[B,-A, x]; A d+Bc I d+Ic ArcTan // Simplify+ Log g 2 d-Ic -ArcTan[1-x] we get x-1 i i + log 3 3 =D -arctan(1-x) 2 x-1 - i 3 3 Combining steps together, yields i x2 -4+i(x+1) 1 log =D arctan x3 -x 2 -3x+5 - arctan(1-x) 2 x2 -4-i(x+1) 3 and 2 x +2x+4 1 3 2 dx= arctan x -x -3x+5 - arctan(1-x) +C x4 -7x 2 +2x+ 17 3.