integrate $\int_0^\pi\frac {1-\alpha^2}{1-2\alpha\cos x +\alpha^2} dx$

I am stuck at this example from Wikipedia on differentiation under the integral sign.

$$\int_0^\pi\frac {1-\alpha^2}{1-2\alpha\cos x + \alpha^2} dx$$

Any help?

Edits:

1. The last term in the denominator is changed to $\alpha^2$.
2. This integral appears in example 3 of the Wikipedia article https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign.
• Are you sure it is the right integral? It would make much more sense if the denominator was $1-2\alpha\cos(x)\color{red}{+\alpha^2}$. Dec 27, 2015 at 1:17
• Please unblock it. Jack D'Aurizio's comment corrects the typo. This integral appears in the Wikipedia article en.wikipedia.org/wiki/Differentiation_under_the_integral_sign in its example 3. With the typo corrected, the integral can be solved. Dec 27, 2015 at 3:18

Let us write the integrand as $$g(x) = \frac{1 - \alpha^2}{1 - 2\alpha\cos x + \alpha^2}$$ We first write $\cos x = \cos^2(x/2) - \sin^2(x/2)$, $1 = \cos^2(x/2) + \sin^2(x/2)$ and $\alpha^2 = \alpha^2(\cos^2(x/2) + \sin^2(x/2))$ to get $$g(x) = \frac{(1 - \alpha^2)\sec^2(x/2)}{(1 - \alpha)^2 + (1 + \alpha)^2\tan^2(x/2)}$$ which is same as $$g(x) = 2\frac{1 + \alpha}{2}\frac{(1 - \alpha)\sec^2(x/2)}{(1 - \alpha)^2 + (1 + \alpha)^2\tan^2(x/2)}$$ or, $$g(x) = 2\frac{1}{1 + \frac{(1 + \alpha)^2}{(1 - \alpha)^2}\tan^2(x/2)}\frac{1 + \alpha}{1 - \alpha}\sec^2(x/2)\frac{1}{2}$$ or $$g(x) = 2 \frac{1}{1 + \left(\frac{1 + \alpha}{1 - \alpha}\right)^2\tan^2\left(\frac{x}{2}\right)}\frac{d}{dx}\left[\frac{1 + \alpha}{1 - \alpha}\tan\left(\frac{x}{2}\right)\right]$$ Therefore, $$\int g(x)dx = 2\tan^{-1}\left(\frac{1 + \alpha}{1 - \alpha}\tan\left(\frac{x}{2}\right)\right) + c,$$ where $c$ is a constant of integration. Therefore, $$\int_0^\pi \frac{1 - \alpha^2}{1 - 2\alpha\cos x + \alpha^2} dx = \begin{cases} \pi & |\alpha| < 1 \\ -\pi & |\alpha| > 1 \end{cases}$$

This is a typical example which can be solved by the residue theorem:

We have that (with $z=e^{ix}$)

\begin{align*} I&= \int_0^\pi\frac {1-\alpha^2}{1-2\alpha\cos x + \alpha^2} dx \\ &=\frac12 \int_{-\pi}^\pi\frac {1-\alpha^2}{1-2\alpha\cos x + \alpha^2} dx \\ &= \frac1{2i} \oint_{|z|=1} \!\frac{dz}z\, \frac{1-\alpha^2}{1 + \alpha^2 -\alpha (z+z^{-1})} \end{align*}

The denominator is given by $$d= z (1+\alpha^2) -\alpha (1+ z^2)$$ and has the poles $z_1=\alpha$ and $z_2=1/\alpha$.

(1) For $|\alpha|<1$ the pole at $z=z_1$ contributes, and we obtain $$I = \pi \mathop{\rm Res}_{z=\alpha} \frac{1-\alpha^2}{d} = \pi.$$

(2) For $|\alpha|>1$ the pole at $z=z_2$ contributes, and we obtain $$I = \pi \mathop{\rm Res}_{z=1/\alpha} \frac{1-\alpha^2}{d} = -\pi.$$

• Residue theorem (almost always) gives an elegant solution. Thanks. Dec 27, 2015 at 7:31

If $|\alpha| < 1$ then we know that $$1 + 2\alpha\cos x + 2\alpha^{2}\cos 2x + \cdots = \frac{1 - \alpha^{2}}{1 - 2\alpha\cos x + \alpha^{2}}\tag{1}$$ Now integrating the above identity with respect to $x$ on interval $[0, \pi]$ we get the desired integral as $\pi$ (because $\int_{0}^{\pi}\cos nx \,dx = 0$). If $|\alpha| > 1$ then we note that $$\frac{1 - \alpha^{2}}{1 - 2\alpha\cos x + \alpha^{2}} = - \dfrac{1 - \dfrac{1}{\alpha^{2}}}{1 - \dfrac{2}{\alpha}\cos x + \dfrac{1}{\alpha^{2}}}$$ and since $|1/\alpha| < 1$ we see that the desired integral is equal to $-\pi$.

i guess, to the case $\vert{\alpha}\vert=1$, your result is different (maybe zero). because, under this assumption, only when $cosx=\frac{a^2+1}{2a}$, your integration will not be zero. so we can compute as below :

$\int_{0}^{\pi}\frac{1-\alpha^2}{1-2\alpha{cosx}+\alpha^2}dx=$$\int_{0}^{\pi}\frac{2(1-\alpha{cosx})}{1-2\alpha{cosx}+\alpha^2}dx pick y=\alpha{cosx} and you can see it : \int_{0}^{\pi}\frac{2(1-\alpha{cosx})}{1-2\alpha{cosx}+\alpha^2}dx=$$\int_{0}^{\pi}\frac{2(1-y)}{1-2y+\alpha^2}dx=$$\int_{-1}^{1}\frac{1}{\sqrt{1-y^2}}dy=$$\alpha{\theta\vert_{-\pi/2}^{\pi/2}}=\vert{\pi}\vert$

but if you multiple $\vert{\pi}\vert$ with $\epsilon$, the result will be zero too!