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]



Integral in the Lamb shift calculation – fourth order
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.


