Variational perturbation theory

Estimate the value of the integral from zero to infinity of exp(-x^2)dx by considering the perturbative expansion of the integral from zero to infinity of exp(- lambda x - p x^2) where p is the small expansion parameter. Perform this expansion to second order in p. The desired integral is then obtained for p = 1 in the limit of lamda to zero, but since we have an expansion in powers of p/lambda^2, this amounts to an infinitely large perturbation, so there seems to be no hope of getting to an accurate estimate of the integral, let alone with a perturbative result to only second order.

However, we can exploit the fact that the lambda x term can be split into a perturbative part and the exact answer should no depend on any such splitting. However, the perturbative result will depend on how we split things up, one then invokes the principle of minimum sensitivity to get to an optimal choice. This works as follows. Put lambda = L + p u where L and u are arbitrary parameters. Re-expand the expression to second order in p. Then eliminate the parameter u by putting
u = (lambda - L)/p. We then end up with a series in p where both lambda and L appear.

In the series it is now possible to set lambda = 0 and we can also set p = 1. We now need to choose L such that the expression is the least sensitive to L, we can do this by equating the derivative equal to zero. It is convenient to work with x = 1/L. Compute the two complex solutions with positive real part, we'll refer to them as s and t.

We can use both solutions to get to a better solution as follows. Denoting the result as function of x as f(x), consider the function g(x), defined as:

g(x) = [f(x) + a f(r x)]/(1+a)

where r = t/s is the ratio of the two solutions. Then g(x) being a weighted average of the result for two parameters should in principle also be suitable to estimate the result. By construction, the derivative of g(x) at x = s is zero, independent of the parameter a. Then choose the parameter a such that the second derivative of g(x) w.r.t. x becomes zero. Show that this yields a much better result than the original estimate.