# Evaluating $\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi$ when $|\rho|<1$

Let $$(X,Y)$$ have a bivariate normal density centered at the origin with $$\mathbb{E}(X^2)=\mathbb{E}(Y^2)=1$$ and $$\mathbb{E}(XY)=\rho$$ with $$|\rho|<1$$. In polar coordinates, $$(X,Y)$$ becomes $$(R,\Phi)$$ where $$X=R\cos\Phi$$ and $$Y=R\sin\Phi$$ . Then it can be shown that $$\Phi$$ has a density given by $$f_{\Phi}(\phi)=\frac{\sqrt{1-\rho^2}}{2\pi(1-\rho\sin2\phi)},\quad 0<\phi<2\pi$$

But I am having difficulty proving that this is indeed a density, more specifically the fact that $$\int_0^{2\pi}f_{\Phi}(\phi)\,\mathrm{d}\phi=1$$ Applying the usual substitution $$t=\tan\phi$$ in the integral $$\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi$$ seems to be misleading. Should there be any change in the integration limits for proving that this a density. If so, why?

Another related integral is $$I=\int_0^{\pi/2}\frac{\mathrm{d}\theta}{1-\rho\sin{2\theta}}=\frac{\pi-\arccos\rho}{\sqrt{1-\rho^2}}\,,$$ which arises while calculating $$\mathbb{P}(X>0,Y>0)$$ for the above distribution once I transform to polar coordinates.

Let $$z=e^{i\phi}$$ and then $$d\phi=\frac{1}{iz}dz$$. So $$\sin\phi=\frac{1}{2i}(z-\frac1z)$$ $$\begin{eqnarray} &&\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi\\ &=&\int_0^{2\pi}\frac{1}{1-\rho \sin{\phi }}\,\mathrm{d}\phi\\ &=&\int_{|z|=1}\frac{1}{1-\rho \frac{1}{2i}(z-\frac1z)}\,\frac{1}{iz}\mathrm{d}z\\ &=&\int_{|z|=1}\frac{2}{2iz-\rho (z^2-1)}\,\mathrm{d}z\\ &=&-\frac{2}{\rho}\int_{|z|=1}\frac{1}{z^2-\frac{2i}\rho z-1}\,\mathrm{d}z\\ &=&-\frac{2}{\rho}\int_{|z|=1}\frac{1}{(z-z_1)(z-z_2)}\,\mathrm{d}z\\ \end{eqnarray}$$ where $$z_{1,2}=\frac{i}{\rho}\pm \frac{\sqrt{1-\rho^2}i}{\rho}$$ are the two root of $$z^2-\frac{2i}\rho z-1=0$$ and only $$z_2$$ is inside $$|z|=1$$. So $$\begin{eqnarray} &&\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi\\ &=&-\frac{2}{\rho}\int_{|z|=1}\frac{1}{(z-z_1)(z-z_2)}\,\mathrm{d}z\\ &=&-\frac{2}{\rho}2\pi i\frac{1}{z_2-z_1}\\ &=&\frac{2\pi}{\sqrt{1-\rho^2}}. \end{eqnarray}$$

Enforcing the substitution $2\phi \to \phi$ and exploiting the $2\pi$-periodicity of the integrand reveals that

\begin{align} \int_0^{2\pi}\frac{1}{1-\rho\sin(2\phi)}\,d\phi&=\frac12\int_0^{4\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi\\\\ &=\frac12 \left(\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi+\int_{2\pi}^{4\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \right)\\\\ &=\frac12 \left(\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi+\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \right)\\\\ &=\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \tag 1 \end{align}

Next, we exploit the $2\pi$-periodicity and the oddness of the sine function in $(1)$ to obtain

\begin{align} \int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi& =2\int_0^\pi \frac{1}{1-\rho^2\sin^2(\phi)}\,d\phi\\\\ &=4\int_0^\pi \frac{1}{(2-\rho^2)+\rho^2\cos(\phi)}\,d\phi\tag2 \end{align}

Applying the tangent half-angle substitution $t=\tan(\phi/2)$ to the integral on the right-hand side of $(2)$ yields

\begin{align} 4\int_0^\pi \frac{1}{(2-\rho^2)+\rho^2\cos(\phi)}\,d\phi&=4\int_0^\infty \frac{1}{(2-\rho^2)+\rho^2 \frac{1-t^2}{1+t^2}}\,\frac{2}{1+t^2}\,dt\\\\ &=4\int_0^\infty \frac{1}{1+(1-\rho^2)t^2}\,dt\\\\ &=\frac{2\pi}{\sqrt{1-\rho^2}} \end{align}

• Could you shed some light on why does using the substitution $\tan \phi=t$ straightaway makes the domain of integration go from $0$ to $0$. Why can't it be applied directly? Sep 23, 2017 at 3:22
• Sure. Note that the tangent function is $\pi$-periodic. Sep 23, 2017 at 3:23

\begin{align} t & = \tan\varphi \\[10pt] dt & = \sec^2\varphi\,d\varphi = (1+\tan^2\varphi)\,d\varphi = (1+t^2)\,d\varphi, \\[10pt] \text{so } \frac{dt}{1+t^2} & = d\varphi \\[10pt] \text{and } \sin(2\varphi) & = 2\sin\varphi\cos\varphi = 2\sin(\arctan t)\cos(\arctan t) \\[10pt] & \phantom{{}= 2\sin\varphi\cos\varphi} = 2\frac{t}{\sqrt{1+t^2}} \cdot\frac{1}{\sqrt{1+t^2}} = \frac{2t}{1+t^2} \end{align} Then we have $$\int_0^{2\pi} \frac{d\varphi}{1 - \rho\sin(2\varphi)} = \left(\int_{-\infty}^\infty + \int_{-\infty}^\infty \right) \frac{\left( \dfrac{dt}{1+t^2} \right)}{1 - \rho\left(\dfrac{2t}{1+t^2}\right)}$$ What do I mean by that?? Simply that as $\varphi$ goes from $0$ to $2\pi,$ then $(\cos(2\varphi),\sin(2\varphi))$ goes around the circle twice, and each revolution causes $t$ to run through the whole real line once. We have a function of period $\pi$ integrated over the interval from $0$ to $2\pi.$ Thus we have $$2 \int_{-\infty}^\infty \frac{\left( \dfrac{dt}{1+t^2} \right)}{1 - \rho\left(\dfrac{2t}{1+t^2}\right)} = 2\int_{-\infty}^\infty \frac{dt}{1+t^2 -2\rho t}.$$ Now complete the square: $$t^2 - 2\rho t + 1 = (t^2 - 2\rho t + \rho^2) + 1 - \rho^2 = (t-\rho)^2 + 1-\rho^2.$$ We would like a quadratic polynomial whose constant term is $1,$ so this becomes \begin{align} & (t-\rho)^2 + 1-\rho^2 = (1-\rho^2)\left( \frac{(t-\rho)^2}{1-\rho^2} + 1 \right) \\[10pt] = {} & (1-\rho^2) \left( \left( \frac{t-\rho}{\sqrt{1-\rho^2}} \right)^2 + 1 \right) \\[10pt] = {} & (1-\rho^2) (u^2 + 1) \\[10pt] \text{so } \sqrt{1-\rho^2}\,\,du & = dt. \end{align} Our integral becomes \begin{align} 2\sqrt{1-\rho^2}\int_{-\infty}^\infty \frac{du}{(1-\rho^2) (u^2+1)} = \frac{2\pi}{\sqrt{1-\rho^2}}. \end{align}

• @StubbornAtom : Could be. But now that's what you've got. Sep 22, 2017 at 18:14

\begin{align} &\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,{d}\phi\\ =& \ \frac1{\sqrt{1-\rho^2}} \bigg[\phi+\cot^{-1}\bigg(\tan2\phi-\frac{1+\sqrt{1-\rho^2}}{\rho}\sec 2\phi \bigg) \bigg]_0^{2\pi} =\frac{2\pi}{\sqrt{1-\rho^2}} \end{align}