# [SOLVED]\int P(r,\theta)

#### dwsmith

##### Well-known member
\begin{alignat}{3}
\int_{-\pi}^{\pi}\frac{1-r^2}{1-2r\cos\theta + r^2}d\theta & = & (2-2r^2)\int_{0}^{\pi}\frac{1}{1-2r\cos\theta + r^2}d\theta
\end{alignat}
We can do the above since the Poisson kernel is even. Wolfram says to make some trig subs which are easily doable but is there a way to integrate in another fashion.
We can use Complex Integration, Residue Theory, or other technique. I would never think of the substitution Wolfram gave so I would like to find a way to do this that is understandable.

#### chisigma

##### Well-known member
\begin{alignat}{3}
\int_{-\pi}^{\pi}\frac{1-r^2}{1-2r\cos\theta + r^2}d\theta & = & (2-2r^2)\int_{0}^{\pi}\frac{1}{1-2r\cos\theta + r^2}d\theta
\end{alignat}
We can do the above since the Poisson kernel is even. Wolfram says to make some trig subs which are easily doable but is there a way to integrate in another fashion.
We can use Complex Integration, Residue Theory, or other technique. I would never think of the substitution Wolfram gave so I would like to find a way to do this that is understandable.
Try using the substitution I suggested in...

http://www.mathhelpboards.com/f10/defeinite-integral-2038/

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
I saw that is what Mathematica said. There isn't another way to do it?
Ok!... first we write the integral in the following form...

$\displaystyle J(r)= \alpha(r)\ \int_{- \pi}^{\pi} \frac{d \theta}{\cos \theta + \gamma (r)}$ (1)

... so that the problem is in some sense 'simplified' to the solution of the integral...

$\displaystyle I(r)= \int_{- \pi}^{\pi} \frac{d \theta}{\cos \theta + \gamma}$ (2)

Now we set $z=e^{i \theta}$ so that is $\displaystyle \cos \theta= \frac{e^{i \theta}+e^{-i \theta}}{2} = \frac{z+z^{-1}}{2}$ and $\displaystyle dz= i\ e^{i \theta} d \theta = i\ z\ d \theta$ and Your integral becomes...

$\displaystyle I= -i \int_{C} \frac{2\ dz}{z^{2}+2\ \gamma\ z +1} = -i \int_{C} f(z)\ dz$ (3)

... where C means 'unit circle'. The poles of the function to be integrated are...

$\displaystyle p_{1}=- \gamma - \sqrt{\gamma^2 -1}\ ,\ p_{2}=- \gamma + \sqrt{\gamma^2 -1}$ (4)

... and their residues are...

$\displaystyle r_{1}= \lim_{z \rightarrow p_{1}} (z-p_{1})\ f(z)\ ,\ r_{2}= \lim_{z \rightarrow p_{2}} (z-p_{2})\ f(z)$ (5)

... and, if neither $p_{1}$ nor $p_{2}$ is on the unit circle [otherwise the integral diverges...] is...

$\displaystyle I= -2\ \pi \sum_{j} r_{j}$ (6)

... where the contribution is of the poles inside the unit circle...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
So $\alpha(r) = 2 - 2r^2$?

Also, what was the substitution for $1 - 2r\cos\theta + r^2 = \cos\theta + \gamma(r)$?

#### chisigma

##### Well-known member
So $\alpha(r) = 2 - 2r^2$?

Also, what was the substitution for $1 - 2r\cos\theta + r^2 = \cos\theta + \gamma(r)$?
Is [without errors of me...] $\displaystyle \alpha(r)= \frac{r^{2}-1}{2\ r}$ and $\displaystyle \gamma(r)= \frac{r^{2}+1}{2\ r}$ ...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
Is [without errors of me...] $\displaystyle \alpha(r)= \frac{r^{2}-1}{2\ r}$ and $\displaystyle \gamma(r)= \frac{r^{2}+1}{2\ r}$ ...

Kind regards

$\chi$ $\sigma$
$$\frac{r^2-1}{2r}\left[\frac{1}{\cos\theta -\left(\frac{1 + r^2}{2r}\right)}\right]$$
so shouldn't it be
$$\frac{1}{\cos\theta - \gamma(r)}$$

#### chisigma

##### Well-known member
$$\frac{r^2-1}{2r}\left[\frac{1}{\cos\theta -\left(\frac{1 + r^2}{2r}\right)}\right]$$
so shouldn't it be
$$\frac{1}{\cos\theta - \gamma(r)}$$
Right!... is $\displaystyle \alpha(r)= \frac{r^{2}-1}{2\ r}$ and $\displaystyle \gamma(r)= -\frac{r^{2}+1}{2\ r}$...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
$\displaystyle p_{1}=- \gamma - \sqrt{\gamma^2 -1}\ ,\ p_{2}=- \gamma + \sqrt{\gamma^2 -1}$ (4)
How do we determine which pole is in the upper half since $\gamma$ is a function of $r$?

#### chisigma

##### Well-known member
How do we determine which pole is in the upper half since $\gamma$ is a function of $r$?
... finding for what r is $|p_{1}(r)|<1$ and $|p_{2}(r)|<1$...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
... finding for what r is $|p_{1}(r)|<1$ and $|p_{2}(r)|<1$...

Kind regards

$\chi$ $\sigma$
$$p_1(r) = \frac{r^2 + 1}{2r}-\sqrt{\left(\frac{r^2 + 1}{2r}\right)^2-1}$$
By the absolute value bars, you are talking about the modulus correct? If so, we have
$$\frac{r^2 + 1}{2r}-\sqrt{\left(\frac{r^2 + 1}{2r}\right)^2-1} < 1$$
Multiplying by the conjugate has
$$1 < \frac{r^2 + 1}{2r}+\sqrt{\left(\frac{r^2 + 1}{2r}\right)^2-1}$$
It just seems circular trying to solve.

#### dwsmith

##### Well-known member
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\left(\frac{r^2 + 1}{2r}\right)$.
$$\frac{r^2 - 1}{2r}\left[\frac{1}{\cos\theta - \frac{1}{2r} - \frac{r^2}{2r}}\right]$$
Then we have
\begin{alignat}{3}
\end{alignat}
Let $z = e^{i\theta}$.
Then $d\theta = -iz^{-1}dz$.
Since $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$, $\cos\theta = \frac{z + z^{-1}}{2}$.
The integral in (1) becomes
$$\frac{-i\alpha}{\pi}\int_C\frac{1}{z^2 + 2z\gamma + 1}dz$$
where $C$ is the contour oriented counterclockwise and with simple poles at $z = -\gamma\pm\sqrt{\gamma^2 - 1}$.
The poles are simple since if we let $g(z) = z^2 + 2z\gamma + 1$, $g'(-\gamma\pm\sqrt{\gamma^2 - 1})\neq 0$ for $0\leq r < 1$.
Let $f(z) = \frac{1}{z^2 + 2z\gamma + 1}$.
Then
$$\int_C\frac{1}{z^2 + 2z\gamma + 1}dz = 2\pi i\sum\text{Res}_{z = z_j}f(z).$$
$z = -\gamma + \sqrt{\gamma^2 - 1}$ is the only pole in $|z| < 1$ in the upper half plane for $0\leq r < 1$.
\begin{alignat*}{3}
2\pi i\sum \text{Res}_{z = z_j}f(z) & = & 2\pi i\left[\frac{1}{z + \gamma - \sqrt{\gamma^2 - 1}}\right]\\
& = & 2\pi i\left[\frac{1}{-\gamma + \sqrt{\gamma^2 - 1} + \gamma + \sqrt{\gamma^2 - 1}}\right]\\
& = & 2\pi i\left[\frac{1}{2\sqrt{\gamma^2 - 1}}\right]\\
& = & \pi i\left[\frac{1}{\sqrt{\left(\frac{r^2 + 1}{2r}\right)^2 - 1}}\right]
\end{alignat*}

\begin{alignat}{3}
\frac{-i\alpha}{\pi}\left(\pi i\left[\frac{1}{\sqrt{\left(\frac{r^2 + 1}{2r}\right)^2 - 1}}\right]\right) & = & \frac{r^2 - 1}{2r}\left[\frac{1}{\sqrt{\left(\frac{r^2 + 1}{2r}\right)^2 - 1}}\right]\\
& = & \frac{r^2 - 1}{\sqrt{(r^2+1)^2-4r^2}}\\
& = & \frac{r^2-1}{\sqrt{(r^2-1)^2}} = 1
\end{alignat}

Last edited:

#### dwsmith

##### Well-known member
I tried the other method but it isn't working out.

What went wrong?

Let $u = \tan\frac{\theta}{2}$.
Then $du = \sec^2\frac{\theta}{2}d\theta$.
Using the trig identities $\sin\theta = 2\sin\frac{\theta}{2} \cos\frac{\theta}{2}$ and $\cos\theta = \cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2}$, we have
$$\sin\theta = \frac{2u}{u^2 + 1},\quad\quad\cos\theta = \frac{1 - u^2}{u^2 + 1},\quad\text{and}\quad d\theta = \frac{2du}{u^2 + 1}.$$
We can now re-write the integral as
\begin{alignat*}{3}
\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}d\theta & = & \frac{1 - r^2}{\pi}\int_{-\pi}^{\pi}\frac{du}{(u^2 + 1)\left(1 - \frac{2r(1 - u^2)}{u^2 + 1} + r^2\right)}\\
& = & \frac{1 - r^2}{\pi}\int_{-\pi}^{\pi}\frac{du}{r^2u^2 + r^2 + 2ru^2 - 2r + u^2 + 1}\\
& = & \left.\frac{1 - r^2}{\pi}\frac{\tan^{-1}\left[\frac{(r + 1)u}{r - 1}\right]}{(r - 1)(r + 1)}\right|_{-\pi}^{\pi}\\
& = & -\left.\frac{\tan^{-1}\left[\frac{(r + 1)\tan\frac{\theta}{2}}{r - 1}\right]}{\pi}\right|_{-\pi}^{\pi}\\
& = & - \frac{\tan^{-1}\left(\infty\right)}{\pi} + \frac{\tan^{-1}\left(-\infty\right)}{\pi}\\
& = & -\frac{\pi}{2\pi} - \frac{\pi}{2\pi}\\
& = & -1
\end{alignat*}