User:Oriol serra i pujol/sandbox

All the elliptic integrands can be reduced to the biquadrate trinomial, x4+sx2+t as we can see in any calculus book (cfr.Puig Adam. Cálculo Integral 1962  cap. 9) Now we will use against the trinomial, the projective transformation of Moebius: the variable change  z=(ax+b)/(cx+d) that has 2 freedom degrees: we can  choose  2 from this constants:(a,b,c,d) We write the trinomial  √Q=√(x2+A)√(x2+B)  (*)    [A & B are real] And after the transformation, we shall have two factors p & q, both trinomials of second (or first) degree, divided by the squared root of  (cx+d)2. Then the integrand is                                         I= ∫dx·√p(x)·√q(x)/(cx+d)2 When p has a double root r, the factor can come out from squared root and we shall have I= ∫dx(x-r)√q(x)/(cx+d)2. (1) 2ªespecie I= ∫dx/(x-r)√q(x)(cx+d)2. (1) 1ªespecie Solved problem, the squared root of the factor q is now of second degree allways  integrable inclusive in combination with factors, like our case. Every elliptic can be solved (*) q(x) can be of first degree, and the denominator also will be.

especial case: it has been “demonstrated” in papers and texts that the integral   √(z�^3-1)  (1960 Davemport IBM)(Ivorra. obra citada pg 43) as impossible. (Now q(x) is of first degree) ******************  I=∫dz√(z3-1)    z�^3-1=(z-1)(1+z+z^2) the quoted transformation projective from Moebius. we will have  z^2=(x^2+2bx+b^2)/(c^2x^2+2cdx+d^2)    (1+z)=[(1+c)x+(w+1)d]/(cx+d) for adition we ought to have in denominator (cx+d)^2 [(1+c)x+(w+1)d)](cx+d)=(c2+c)x2+x[cd(1+w)+d(1+c)]+d2(1+w)   id  est   (1+z)(cx+d)2       z-1=[(1-c)x+d(w-1)]/(cx+d)   The root of this monomiall  is x=d(w-1)/(c-1)  and we need it would be also root for the other factor

[z^2+(1+z)]/(cx+d)^2=(x^2+2bx+b^2)+(c+c^2)x*2+x[(1+c)d+cd(1+w)]+d^2(w+1) hacemos c=-1   b=wd  d, es la escala. (x^2+2wx+w^2)-x[(w+1)]+(w+1) x^2+x(w-1)+(w+1)=0 the root  has to be  x=(1-w)/2 and is x=(1/2)[1-w+-√(w^2-6w-3)    the radical has to be  zero w=3+-2√3     x=-1-√3 h(-)  - the integrand I=∫dx(x+1+√3)√[x-h(+)]/(1+√3-x)2  the factor  comes ut from radical. the other root  is h(+)     I=∫dx(x+1+√3)√(x+h(+)/(h-x)2   we do  x=t^2-h(+)    dx=tdt    I=∫tdt·(t^2+2√3)/[2+t2]^2          I=∫t^4dt/(2+t^2)^2+2√3∫t^2dt/(2+t^2)^2  t=√2shq             I=∫dqsh^4(q)/ch^3(q)+2√3∫dq·sh^2(q)/ch^3(q)    q=arc.sh[√(x+h)/√2]  wikipedia  solves theese         n=3  m=4 y 2

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