an alebraic integral

A certain algebraic integral PhL 3.26.05 Note that page numbering is turned on in this template. !Syntax Error, Idθ sinθ/* cos-1[/] = π [cos(φ/2) - cos(θ0/2) ] (G.22) = *** cos-1[/] = cos-1[cos(θ0/2)/cos(θ/2)] sinθ = 2sin(θ/2)cos(θ/2) *** so LHS = !Syntax Error, Idθ sin(θ/2)cos(θ/2)/* cos-1[cos(θ0/2)/cos(θ/2)] Let x = cos(θ/2) and a = cos(φ/2) and b = cos(θ0/2). Then dx = - (1/2) sin(θ/2) dθ so get *** LHS = !Syntax Error, I 2 dx x / * cos-1(b/x) = (2) !Syntax Error, I dx x / * sec-1(x/b) Now write x (a2-x2)-1/2 = 2 (-1/2) ∂x (a2-x2)1/2 = - ∂x (a2-x2)1/2 Maple agrees so now LHS = –(2) !Syntax Error, I dx ∂x (a2-x2)1/2 sec-1(x/b) = – (2) [ (a2-x2)1/2 sec-1(x/b)]|ab + (2)!Syntax Error, I dx(a2-x2)1/2 ∂x sec-1(x/b) = (2)!Syntax Error, I dx (a2-x2)1/2 ∂x sec-1(x/b) But we know that ∂x sec-1(x/b) = (b/x) /* (1/b) = (1/x) / = (b/x) / So we then have LHS = (2b) !Syntax Error, I dx x-1 (a2-x2)1/2 (x2-b2)-1/2 Now let y = x2 so dy = 2xdx and get dx/x = dy/(2x2) = dy/(2y). Let a2 = α and b2 = β. *** LHS = (2b)!Syntax Error, I dy/(2y) (α-y)1/2 (y-β)-1/2 = (b) !Syntax Error, I dy/(y) (α-y)1/2 (y-β)-1/2 Now we want to look up this integral I think this has to get reduced to an F function. I cannot find it in GR7. Maple can't do it. But Wolfram can do it and gets π ( - 1) So our result is LHS = (b) π ( a/b - 1) = (π) * (a - b) = π (cos(φ/2) - cos(θ0/2)) So how did Wolfram do this integral? It can do the indefinite integral! Well, I have an idea Let z = (a-y)/(y-b) so the integration range is then 0 to ∞ Maple shows that dy/y = dz (b-a)/[(z+1)(zb+a)] . The integral in blue above becomes (a-b)!Syntax Error, Idz / [(z+1)(zb+a)] = (a-b)/b* !Syntax Error, Idz / [(z+1)(z+a/b)] = ([a/b] - 1) !Syntax Error, Idz / [(z+1)(z+[a/b])] Let c = a/b so integral is then (c-1) !Syntax Error, Idz / [(z+1)(z+c)] Now write 1/[(z+1)(z+c) = (c-1)-1 [ 1/(z+1) - 1/(z+c) ] Then our integral becomes !Syntax Error, Idz / [ 1/(z+1) - 1/(z+c) ] = !Syntax Error, Idz/(z+1) - !Syntax Error, Idz/(z+c) But each term diverges! Suppose we regulate with a cutoff? Maple says Thus, the zero endpoint gives nothing and we have !Syntax Error, Idz/(z+c) = 2 - 2 tan-1 (/) Our difference will then be !Syntax Error, Idz/(z+c) - !Syntax Error, Idz/(z+1) = [2 - 2 tan-1 (/) ] - [2 - 2 tan-1 () = 2 tan-1 () - 2 tan-1 (/) Then we take Λ→∞ and tan-1∞ = π/2 and we have = π - π = π(1-) This we have shown that !Syntax Error, I dy/(y) (α-y)1/2 (y-β)-1/2 = (c-1) !Syntax Error, Idz / [(z+1)(z+c)] = regulate { !Syntax Error, Idz/(z+1) - !Syntax Error, Idz/(z+c) } = π(-1) So here is how this all goes: Start with R = !Syntax Error, Idθ sinθ/* cos-1[/] Let x = cos(θ/2) and a = cos(φ/2) and b = cos(θ0/2) to get R = (2) !Syntax Error, I dx x / * sec-1(x/b) Write x (a2-x2)-1/2 = - ∂x (a2-x2)1/2 and do parts (parts vanish) to get R = (2)!Syntax Error, I dx (a2-x2)1/2 ∂x sec-1(x/b) Compute ∂x sec-1(x/b) = (b/x) / to get R = (2b) !Syntax Error, I dx x-1 (a2-x2)1/2 (x2-b2)-1/2 Let y = x2 to get R = (b) !Syntax Error, I dy/(y) (α-y)1/2 (y-β)-1/2 a2 = α and b2 = β Let z = (a-y)/(y-b) to get R = ([a/b] - 1) !Syntax Error, Idz / [(z+1)(z+[a/b])] Let c = a/b etc etc. OK, this is not really any faster than my original method. ************************** or R = !Syntax Error, I dz /) * cos-1[/] = !Syntax Error, Idz (b-z)-1/2 cos-1[/] a = cosθ0 b = cosφ where the reader is asked to forget about the previous a and b. In this form we can do parts integration to get rid of the painful arc trig function. The "parts" vanish and we are left with R(φ) = 2 !Syntax Error, Idz (b-z)1/2 ∂z sec-1[/] = !Syntax Error, Idz (b-z)1/2 / [(1+z) ] let z = y+a: = !Syntax Error, Idy ([b-a]-y)1/2 (y + 1+a)-1 y-1/2 let s = y-1; = (b-a)1/2 (1+a)-1/2 !Syntax Error, I ds s-1 ( s - [b-a]-1)1/2(s + [1+a]-1)-1 = (π/2) (b-a) (b+1)1/2 (a+1)-1 { F(1,3/2; 2; -(b-a)/(1+a))} // see below = (π/2) (b-a) (b+1)1/2 (a+1)-1 { 2(a+1) [ - ] / [(b-a) ] } = π [ - ] = π [cos(φ/2) - cos(θ0/2) ] (G.20) where we used corrected GR7 page 317 3.197.2, !Syntax Error, Ids s-λ (s-u)μ-1 (s+β)ν = u-λ (β+u)μ+ν B(λ-μ-ν,μ) F(λ,μ; λ-ν; -β/u) (G.21) So we have shown that !Syntax Error, Idθ sinθ/* cos-1[/] = π [cos(φ/2) - cos(θ0/2) ] (G.22)