Integral in the Lamb shift calculation – fourth order

Question

I was trying to solve some integrals that appear in quantum electrodynamics but I was not able to do it on my own.

$$1/6\int_0^1 \int_0^1 { u^3 z^2(1-z^2/3) \over [u^2(1-z^2)+4(1-u)]}dudz $$

I know the answer should be $$(\pi^2 / 18) - (115 / 216)$$

Can anyone help me solve this or at least point me to a book that might cover those double integrals? I would realy appreciate it. In attachment you have an image of the paper this came from. Of course I have problems with the other integrals too. But I chose to start with this one.

Thank you.

[Extract of paper by Soto, 1970 - Calculation of the Slope at q2=0 of the Dirac Form Factor for the Electron Vertex in Fourth Order][1]

Answer 1

The double integral in question is $$I:=\int_0^1 du\, J(u),$$ where $$J(u):=\int_0^1 dz\,{ u^3 z^2(1-z^2/3) \over {u^2(1-z^2)+4(1-u)}} \\ \text{[which is a standard integral, with a denominator of the integrand quadratic in $z$]} \\ =\frac{3 \left(u^3-6 u+4\right) \ln(1-u)+u \left(-5 u^2-12 u+12\right)}{9 u^2} \\ =J_1(u)-J_2(u),$$ $$J_1(u):=\frac{3 \left(u^3-6 u+4\right)(-u)+u \left(-5 u^2-12 u+12\right)}{9 u^2}=\frac19\, (6 - 5 u - 3 u^2),$$ $$J_2(u):=\sum_{k=2}^\infty \frac{3 \left(u^3-6 u+4\right) u^k/k}{9 u^2}; $$ here we used the Maclaurin series for $\ln(1-u)$.

The integral $\int_0^1 du\, J_1(u)$ is very easy, and $$-\int_0^1 du\, J_2(u)=\sum_{k=2}^\infty \Big(\frac{2}{k^2}+\frac{7}{6 k}+\frac{1}{6 (k+2)}-\frac{4}{3 (k-1)}\Big).$$ Next, $\sum_{k=2}^\infty \frac2{k^2}=2(\pi^2/6-1)$, and the sum $$\sum_{k=2}^\infty \Big(\frac{7}{6 k}+\frac{1}{6 (k+2)}-\frac{4}{3 (k-1)}\Big) =-\frac{53}{36}$$ can be found either by telescoping or using the fact that $\sum_{k=1}^n\frac1k-\ln n$ converges as $n\to\infty$ to Euler's gamma constant $\gamma=0.577\ldots$ (which latter will get canceled, because $\frac76+\frac16-\frac43=0$).

Collecting the pieces, we get the result.