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Very complicated integral

I'm looking for a solution to the following integral:

int (1/sqrt[1 - x^2 + a (x^6 - x^12)]) dx

where a is a parameter.

Could you help me to find a solution?

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William S. | Experienced scientist, mathematician and instructor - WilliamExperienced scientist, mathematician and...
4.4 4.4 (10 lesson ratings) (10)
If I'm interpreting this correctly, you are looking for the indefinite integral of
{sqrt[(1-x2 + a(x6-x12)]}-1 or [1-x2 + a(x6-x12]-1/2
Am I correct?


Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
Try substitute x=sin(t); The integral becomes
This integral can be evaluated using the following expansion:
(1-x)=∑0(2k-1)!!/(2k)!!*xk, summed over k. This series converges for x<1.
Expression f(t)=a*sin6(t)(1+sin2(t)+sin4(t)) needs to be less than 1. This puts restrictions on the a parameter. [f(t)]k is integrable, however, integrals will be very ugly looking. For example,
  213 x      337 Sin[2 x]        41 Sin[4 x]      31 Sin[6 x]
{-------  - ---------------     + -------------  -  -------------   +
  256              12                    256                 1024

7 Sin[8 x]        Sin[10 x]
------------   -  ----------- }
 2048                5120


As an addition to Kirill's method: the substitution x=sin(t) only works for |x|≤1. At x=±1, the integral becomes improper and needs to be split. For |x|≥1, you could use another trig substitution, perhaps x=sec(t).
In fact, I forgot a parameter in the formula, so the integral becomes:
int (1/sqrt[1 - x^2 + a (2(bx)^6 - (bx)^12)]) dx

I'm sorry for this error.
So now your potential is U(x) = a (2(bx)6-(bx)12). You should also mention that a and b are very small positive numbers. The equilibrium point now is at x0=1/b. Since x=1/r, b is the (stable) equilibrium distance between the interacting atoms/molecules, usually a few Angstroms. With U(1/b)=a, U'(1/b)=0, and U''(1/b)=-72ab, you get the quadratic approximation U(x)≈a(1-36 b(x-1/b)²) and can find the small frequency oscillations about the equilibrium.
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
This integral represents the integral curves φ(r) (in polar coordinates, with x=1/r) for the famous 12-6 Lennard-Jones potential for atomic interactions. It is usually treated quantum-mechanically, not as a classical central potential problem.
You should graph the potential, U= a(x6-x12), and check it has a stable equilibrium at x0=2-1/6, corresponding to circular orbits.Deduce the nature of the other orbits (elliptic, parabolic, hyperbolic) qualitatively from your graph. Then do a Taylor series expansion of U about x0 up to order x2, so you will get a harmonic oscillator potential. Substitute this into your integral and get the well-know harmonic oscillator solutions. I doubt there exists an analytic expression for the unapproximated potential.


I think an analytical approach is more significant from the physical point of view, especially highlighting the phenomenons of orbiting and spiraling of the particles
If you find an analytic solution, you should publish it in JMC (Journal of Mathematical Chemistry)!
Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (638 lesson ratings) (638)
The Mathematica Integrator says that the integral of this function probably does not exist.