Quadruple Integral in the Lamb Shift

[Francisco Alegria]

TL;DR Summary

Computation of a quadruple integral that comes up when computing the fourth order contribuition to the Lamb Shift in energy of the electron orbiltals - Self energy part

The analytical computation of the shift in energy level of electrons in atoms due to quantum electrodynamics is carried out using perturbation theory. In particular, the fourth-order contribution is given in five different terms. One of them, usually called "Electron Self Energy", leads to seven different quadruple integrals. I do not know how to compute any of them on my own.

I ask anyone for some assistance in computing one of the easiest ones (with what I learn from you, I hope to be able to do the other ones).
Here it is:

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{w(w-1)(1-u^2)v^2}{v(1-u)z+(1-w)u}dudzdvdw$$

The result reported in M. F. Soto, "Calculation of the Slope at q^2=0 of the Dirac Form Factor for the Electron Vertex in Fourth Order", Physical Review A, vol. 2, no. 3, pp. 734-758, September 1970, eq. (A7) and (A8) is $\pi^2/120-5/32$.

Which integral should I do first?

 

[member38644]

First, we can rewrite the integral as follows:

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{w(w-1)(1-u^2)v^2}{v(1-u)z+(1-w)u}dudzdvdw = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{w(w-1)(1-u^2)v}{(1-u)z+(1-w)u}dudzdvdw$$

Next, we can use the substitution $x = (1-u)z + (1-w)u$ to simplify the integral. This substitution allows us to rewrite the integral as follows:

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{w(w-1)(1-u^2)v}{x}dudzdvdw$$

To compute this integral, we can use the method of partial fractions. First, we can factor the numerator as follows:

$$w(w-1)(1-u^2)v = w(u+1)(u-1)v = w(u^2-1)v$$

Next, we can rewrite the integral as follows:

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{w(u^2-1)v}{x}dudzdvdw$$

We can now split the integral into four separate integrals, each with respect to one of the variables:

$$\int_{0}^{1}\frac{w}{x}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(u^2-1)v\,dudzdvw$$

$$+\int_{0}^{1}\frac{u^2-1}{x}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}wv\,dudzdvw$$

$$+\int_{0}^{1}\frac{v}{x}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}w(u^2-