Quadruple Integral in the Lamb Shift

In summary, the conversation discusses the analytical computation of the shift in energy level of electrons in atoms due to quantum electrodynamics using perturbation theory. The fourth-order contribution is given in five different terms, one of which is the "Electron Self Energy" term that involves seven quadruple integrals. The speaker is seeking assistance in computing one of the integrals and hopes to use the knowledge gained to compute the others. The specific integral mentioned is reported in a publication and the result is provided. The speaker asks for guidance on which integral to compute first.
  • #1
Francisco Alegria
2
0
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?
 
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  • #2


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-
 

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