Solving Elliptic Integrals to Understanding y² Polynomials in x

  • Thread starter Thread starter koolmodee
  • Start date Start date
  • Tags Tags
    Integrals
koolmodee
Messages
51
Reaction score
0
\int(1-x^{4})^{-1/2}dx
and the integral goes over [0,1]

The book I'm reading says this is an elliptic integral, meaning that it is the integral of a rational function of x and y in which y² is a polynomial in x, of degree 3 or 4, having simple roots.

What does the author possibly mean?

What y? Why y² polynomial in x? Why degree 3 0r 4? And yes, why having simple roots?

Could someone give some hints?

thank you
 
Physics news on Phys.org
Actually, I am not familiar with that definition! An "elliptic integral" is a generalization of the kind of integral you get trying to calculate the circumference of an ellipse. In any case, here if you take y2 to be 1- x4, a fourth dergee polynomial with obviously simple roots, then (1- x^4)^{-1/2}= \frac{1}{\sqrt{1- x^4}}= \frac{1}{\sqrt{y^2}}= \frac{1}{y}. a rational function of y.
 
many thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top