Poisson Integral is equal to 1

Question

Show $$ \int_{-\pi}^{\pi}P(r, \theta)d\theta = 1 $$

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Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$. Then $$ \frac{1}{2\pi} \int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta = \frac{\alpha}{2\pi} \int_{-\pi}^{\pi}\frac{1}{\cos(\theta) + \gamma}d\theta $$ where $$ \frac{r^2 - 1}{2r}\frac{1}{\cos(\theta) - \frac{1}{2r} - \frac{r^2}{2r}} = \frac{1 - r^2}{1 - 2r\cos(\theta) + r^2} $$ Next, let $z = e^{i\theta}$. Then $d\theta = \frac{-i}{z}dz$. Since $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$, $\cos(\theta) = \frac{z + z^{-1}}{2}$. $$ = \frac{-i\alpha}{\pi}\int_C\frac{1}{z^2 + 2z\gamma + 1}dz $$ and $C$ is the contour oriented counter clockwise with simple poles at $z = -\gamma\pm\sqrt{\gamma^2 - 1}$. Let $f(z) = \frac{1}{z^2 + 2z\gamma + 1}$. Then $$ = \Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z) \tag{1} $$ The only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$. Then $$ 2\pi i\lim_{z\to z_j}\Bigg[\big(z + \gamma - \sqrt{\gamma^2 - 1}\big) \frac{1}{\big(z_j + \gamma + \sqrt{\gamma^2 - 1}\big) \big(z + \gamma - \sqrt{\gamma^2 - 1}\big)}\Bigg] = \frac{\pi i}{\sqrt{\gamma^2 - 1}} $$ Now, we can substitute $\frac{\pi i}{\sqrt{\gamma^2 - 1}}$ for $2\pi i\sum\text{Res}$ in equation (1). \begin{align*} \Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z) &= \frac{\alpha}{\sqrt{\gamma^2 - 1}}\\ &= \frac{r^2 - 1}{2r\sqrt{\frac{(r^2 + 1)^2}{4r^2} - 1}}\\ &= \frac{r^2 - 1}{\sqrt{(r^2 + 1)^2 - 4r^2}}\\ &= \frac{r^2 - 1}{\sqrt{r^4 - 2r^2 + 1}}\\ &= \frac{(r - 1)(1 + r)}{(r - 1)(r + 1)}\\ &= 1 \end{align*}

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I have been unable to convince myself that the only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$. I know it is the case because if I use the other pole, the integral becomes $-1$

Answer 1

For $\boldsymbol{r\in(-1,1)}$ $$ \begin{align} 1 &=\frac1{2\pi i}\oint\frac1{z-r}\mathrm{d}z\tag{1}\\ &=\frac1{2\pi i}\int_{-\pi}^\pi\frac1{e^{i\theta}-r}\mathrm{d}e^{i\theta}\tag{2}\\ &=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-re^{i\theta}}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{3}\\ &=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-r\cos(\theta)}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{4}\\ &=\frac12+\frac1{2\pi}\int_{-\pi}^\pi\frac{\frac12-\frac12r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{5}\\ 1 &=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{6} \end{align} $$ Explanation:
$(1)$: $\frac1{z-r}$ has residue $1$ at $z=r$ (inside the unit circle)
$(2)$: parametrize the unit circle
$(3)$: multiply integrand by $\frac{e^{-i\theta}-r}{e^{-i\theta}-r}$
$(4)$: take the real part of both sides
$(5)$: add $\frac12$ to and subtract $\frac12$ from the integral
$(6)$: multiply both sides by $2$ and subtract $1$

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For $\boldsymbol{r\not\in(-1,1)}$

If $r^2=1$, the integrand is $0$, so assume $r\not\in[-1,1]$.

If $r\not\in[-1,1]$, then the left side of $(1)$ starts at $0$ and step $(6)$ changes the left side to $-1$.

Alternatively, if $r\not\in[-1,1]$, then $\frac1r\in(-1,1)$, and therefore, $(6)$ says $$ \begin{align} \frac1{2\pi}\int_{-\pi}^\pi\frac{1-r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta &=\frac1{2\pi}\int_{-\pi}^\pi\frac{\frac1{r^2}-1}{\frac1{r^2}-\frac2r\cos(\theta)+1}\mathrm{d}\theta\\ &=-\frac1{2\pi}\int_{-\pi}^\pi\frac{1-\frac1{r^2}}{1-\frac2r\cos(\theta)+\frac1{r^2}}\mathrm{d}\theta\\ &=-1\tag{7} \end{align} $$

Answer 2

For $0 \le r < 1$, $$ \begin{align} \sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta} & =\sum_{n=0}^{\infty}r^{n}e^{in\theta}+\sum_{n=1}^{\infty}r^{n}e^{-in\theta} \\ & =\frac{1}{1-re^{i\theta}}+\left(\frac{1}{1-re^{-i\theta}}-1\right) \\ & = \frac{1}{1-re^{i\theta}}+\frac{re^{-i\theta}}{1-re^{-i\theta}} \\ & = \frac{1-re^{-i\theta}+re^{-i\theta}-r^{2}}{(1-re^{i\theta})(1-re^{-i\theta})} \\ & = \frac{1-r^{2}}{1-2r\cos\theta+r^{2}}. \end{align} $$ Therefore, because $\int_{-\pi}^{\pi}e^{in\theta}\,d\theta=0$ for $n \ne 0$, the following holds for $0 \le r < 1$: $$ \int_{-\pi}^{\pi}\frac{1-r^{2}}{1-2r\cos\theta+r^{2}}\,d\theta= \sum_{n=-\infty}^{\infty}r^{|n|}\int_{-\pi}^{\pi}e^{in\theta}\,d\theta = \int_{-\pi}^{\pi}d\theta=2\pi. $$

Answer 3

Let $z_+ = -\gamma + \sqrt{\gamma^2 - 1}$ and $z_- = -\gamma - \sqrt{\gamma^2 - 1}$. Then \begin{align} z_+ &= \frac{r^2 + 1}{2r} + \sqrt{\frac{r^4 + 2r + 1 - 4r^2}{4r^2}}\\ &= \frac{r^2 + 1}{2r} + \frac{r^2 - 1}{2r}\\ &= \frac{2r^2}{2r}\\ &= r\\ z_- &= \frac{r^2 + 1}{2r} - \sqrt{\frac{r^4 + 2r + 1 - 4r^2}{4r^2}}\\ &= \frac{r^2 + 1}{2r} - \frac{r^2 - 1}{2r}\\ &= \frac{2}{2r}\\ &= \frac{1}{r} \end{align} So for $0<r<1$, only $z_+$ is in the unit circle.

Answer 4

From your definition, $\gamma = -\frac{r^2+1}{2r}$ so $-\gamma>1$ for all positive $r$. So $-\gamma+\sqrt{\gamma^2-1}>-\gamma>1$ so that root cannot possibly lie in $|z|<1$. Are you sure you do not have a sign error somewhere between lines 3 and 4 of your derivation?