Residue theorem in case of discontinuous function

Question

I need to compute the following integral: $$ m=2(1-z^2)\int^\pi_0\frac{g(\theta,x)}{z^2-2z\cos\theta+1}d\theta $$ where $$ g(\theta,x)= \begin{cases} \frac{1}{2\pi}\frac{\sin(\theta/2)}{\sqrt{\sin^2(\theta/2)-x^2}} \ \ \mathrm{if} \ \ \cos(\theta)\leq1-2x^2 \\ 0 \ \ \mathrm{else} \end{cases} $$ I (think I) know the answer is $$ m=\sqrt{\frac{z^2-2z+1}{z^2-2z(1-2x^2)+1}} $$ but I need to make my proof more rigorous as it just stands on intuition for the moment... I have used the residue theorem, considering that the density $g(\theta)$ is an analytic measure on the domain of integration and that the denominator has simple zeros at $e^{\pm i\theta_0}=z$. Taking that $$ \mathrm{Res}\left[\frac{g(\theta,x)}{z^2-2z\cos\theta+1},\theta_0\right]=\frac{g(\theta_0,x)}{(z^2-2z\cos\theta+1)'} $$ directly yields the right answer, if I consider a contour so that the residue is multiplied by $2\pi i$ and not just $\pi i$.

Clearly this is not rigorous as I don't even know what domain of integration would allow me to use the residue theorem as such. Besides, I don't know to which extent my reasoning is wrong due to the discontinuity in $g$. Any help would be much appreciated! Thanks

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} m & \equiv 2\pars{1 -z^{2}}\int_{0}^{\pi}{1 \over 2\pi} \,{\sin\pars{\theta/2} \over \root{\sin^{2}\pars{\theta/2} -x^{2}}}\, {\bracks{\cos\pars{\theta} \leq 1 - 2x^{2}} \over z^{2} - 2z\cos\pars{\theta} + 1}\,\dd\theta \\[1cm] = &\ {2 \over \pi}\pars{1 -z^{2}}\int_{0}^{\pi} \,{\sin\pars{\theta/2} \over \root{1 - \cos^{2}\pars{\theta/2} -x^{2}}}\times \\[5mm] &\ \bracks{2\cos^{2}\pars{\theta \over 2} - 1 \leq 1 - 2x^{2}} {\dd\theta/2 \over z^{2} - 2z\bracks{2\cos^{2}\pars{\theta/2} - 1} + 1} \\[1cm] \stackrel{t\ =\ \cos\pars{\theta/2}}{=}\,\,\, & {2 \over \pi}\pars{1 -z^{2}}\int_{0}^{1} {\bracks{t \leq \root{1 - x^{2}}} \over \root{1 - x^{2} - t^{2}}\pars{z^{2} + 2z + 1 - 4zt^{2}}}\,\dd t \\[5mm] = &\ {2 \over \pi}\pars{1 -z^{2}}\int_{0}^{\root{1 - x^{2}}} {\dd t \over \root{1 - x^{2} - t^{2}}\bracks{\pars{z + 1}^{2} - 4zt^{2}}} \\[5mm] \stackrel{t\ =\ \root{1 - x^{2}}\cos\pars{\phi}}{=}\,\,\, & {2 \over \pi}\pars{1 -z^{2}}\int_{0}^{\pi/2} {\dd\phi \over \pars{z + 1}^{2} - 4z\pars{1 - x^{2}}\cos^{2}\pars{\phi}} \\[5mm] = &\ {2 \over \pi}\pars{1 -z^{2}}\int_{0}^{\pi/2} {\sec^{2}\pars{\phi}\, \dd\phi \over \pars{z + 1}^{2}\sec^{2}\pars{\phi} - 4z\pars{1 - x^{2}}} \\[5mm] = &\ {2 \over \pi}\pars{1 -z^{2}}\int_{0}^{\pi/2} {\sec^{2}\pars{\phi}\, \dd\phi \over \pars{z + 1}^{2}\tan^{2}\pars{\phi} + \pars{z - 1}^{2} + 4zx^{2}} \\[5mm] = &\ {2 \over \pi}\pars{1 -z^{2}}\,{\pi \over 2}\, {1 \over \pars{z + 1}\root{\pars{z - 1}^{2} + 4zx^{2}}} = \bbx{\ds{1 - z \over \root{\pars{z - 1}^{2} + 4zx^{2}}}} \end{align}