Angular integrals used in QED

An interesting approach to these integrals is the use of complex integration. In this case, we can use the fact that with \begin{equation} f(\sin\theta,\cos\theta) = \frac{\sin^2\theta \cos^{2k} \theta}{\beta^2-\cos^2\theta} \end{equation} we have \begin{eqnarray} \int_0^\pi \mathrm{d}\theta \, f(\sin\theta,\cos\theta) &=& \frac{1}{2} \int_0^{2\pi} \mathrm{d}\theta \,f(\sin\theta,\cos\theta) \\ &=& \frac{1}{2} \int_{|z|=1} \frac{\mathrm{d} z}{i z} \, f\left(\frac{z-z^{-1}}{2i},\frac{z+z^{-1}}{2}\right) \\ &=& - \frac{i}{2^{2k+1}} \,\int_{|z|=1} \mathrm{d} z\, \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} (z^4-4 \beta^2 z^2+2z^2+1)} \\ &=& - \frac{i}{2^{2k+1}} \,\int_{|z|=1} \mathrm{d} z\, \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} \Pi_{i=1}^4 (z-z_i)} \, , \end{eqnarray} with $z_i$ the simple poles \begin{eqnarray} z_1 &=& -\beta - \sqrt{\beta^2 - 1} \\ z_2 &=& \beta - \sqrt{\beta^2 - 1} \\ z_3 &=& -\beta + \sqrt{\beta^2 - 1} \\ z_4 &=& \beta + \sqrt{\beta^2 - 1} \,. \end{eqnarray} By looking at the result, seems like the author assumes the range of $\beta$ is such that only $z_2$ and $z_3$ lay inside the unit circle (besides the $(2k+1)$-order pole at the origin). Add the contributions from those 3 poles and you can actually find a formula for arbitrary $k$ with relative ease.

More precisely, if we write \begin{equation} g = \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^{2k+1} \Pi_{i=1}^4 (z-z_i)} \, \end{equation} then we find \begin{eqnarray} \int_0^\pi \mathrm{d}\theta \, f(\sin\theta,\cos\theta) = && \frac{\pi}{2^{2k}} \left\{ \frac{1}{(2k)!} \lim_{z\rightarrow 0} \frac{d^{2k}}{dz^{2k}} \left[z^{2k+1} g\right] + \lim_{z\rightarrow z_2} \left[(z-z_2)g\right] + \lim_{z\rightarrow z_3} \left[(z-z_3)g\right] \right\} \, . \end{eqnarray} Replacing and simplifying gives us \begin{equation} I_k(y) = \frac{\pi}{2^{2k}} \frac{1}{(2k)!} \lim_{z\rightarrow 0} \frac{d^{2k}}{dz^{2k}} \left[ \frac{(z^2-1)^2 (z^2+1)^{2k}}{z^4+z^2\left[2-4\beta^2(y)\right] + 1} \right] - \pi \beta^{2k-1}(y) \sqrt{\beta^2(y) - 1} \, . \end{equation} Finally, when expanding the first term, it turns out this can be further simplified, producing \begin{equation} I_k(y) = - \pi \sum_{m=0}^k \frac{(2m-3)!!}{(2m)!!} \beta^{2k-2m}(y) - \pi \beta^{2k-1}(y) \sqrt{\beta^2(y) - 1} \, , \end{equation} where we take $(-3)!! = -1$, $(-1)!! = 1$ and $0!! = 1$. This also allows us to generalize the relation between successive values of $I_k(y)$, where we have \begin{equation} I_k(y) = - \frac{(2k-3)!!}{(2k)!!} \pi + \beta^{2}(y) I_{k-1}(y) \,. \end{equation}

In hindsight, this result could have been deduced in a couple of lines, by noticing that the difference between the integrands of $I_k(y)$ and $\beta^2(y) I_{k-1}(y)$ simplifies to \begin{equation} \frac{\sin^2\theta \cos^{2k} \theta}{\beta^2(y)-\cos^2\theta} - \beta^2(y) \frac{\sin^2\theta \cos^{2k-2} \theta}{\beta^2(y)-\cos^2\theta} = \cos^{2k} \theta - \cos^{2k-2} \theta \,. \end{equation} Hence \begin{eqnarray} I_k(y) - \beta^2(y) I_{k-1}(y) &=& \int_0^\pi \mathrm{d}\theta \left[ \cos^{2k} \theta - \cos^{2k-2} \theta \right] \\ &=& \pi \left[ \frac{(2k-1)!!}{(2k)!!} - \frac{(2k-3)!!}{(2k-2)!!} \right] \\ &=& - \frac{(2k-3)!!}{(2k)!!} \pi \, , \end{eqnarray} for $k \geq 1$.

For $I_0(y)$, manipulate the fraction using the following: $$I=\int_0^{\pi} \frac{\sin^2{x}+\cos^2{x}-\cos^2{x}+{\beta}^2(y)^2-{\beta}^2(y)^2}{{\beta}^2(y)-\cos^2{x}} \; dx = \int_0^{\pi} \frac{1-{\beta}^2(y)}{{\beta}^2(y)-\cos^2{x}} \; dx$$ $$+ \int_0^{\pi} \frac{{\beta}^2(y)-\cos^2{x}}{{\beta}^2(y)-\cos^2{x}} \; dx$$ $$=\int_0^{\pi} \frac{1-{\beta}^2(y)}{{\beta}^2(y)-\cos^2{x}} \;dx + \pi$$

Now, let $t=\tan{x}$ and the integral simplifies to : $$2 \int_0^{\infty} \frac{1-{\beta}^2(y)}{{\beta}^2(y) \; t^2+\left({\beta}^2(y)-1\right)} \; dt$$

This is easy to solve as it is in the form of $\arctan{u}$: $$2\frac{\sqrt{{\beta}^2(y)-1}}{{\beta}(y)} \arctan{\left(\frac{{\beta}(y) \;t}{\sqrt{{\beta}^2(y)-1}}\right)} \big \rvert_0^{\infty}= \pi \frac{\sqrt{{\beta}^2(y)-1}}{b{\beta}(y)}$$.

Now add this to the $\pi$ from line 2: $$\boxed{I_0(y)=\pi\left(1-\sqrt{1-\frac{1}{{\beta}^2(y)}}\; \right)}$$

Similarly, for $I_1(y)$, manipulate the fraction using the following: $$I_1(y)=\int_0^{\pi} \frac{\sin^2{x}\left(1-\sin^2{x}\right)}{{\beta}^2(y)-\cos^2{x}} \; dx= I_0(y)-\int_0^{\pi} \frac{\sin^4{x}}{{\beta}^2(y)-\cos^2{x}} \; dx$$

Try finishing it.