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Very complicated integral

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4 Answers

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?

Comments

Try substitute x=sin(t); The integral becomes
 
∫dt/√[1+a*sin6(t)*(1+sin2(t)+sin4(t))];
 
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,
 
∫f(t)dt=a*
  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

Comments

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.
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.

Comments

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)!