5

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.
Asaf Karagila
  • 393,674
Rodrigo
  • 7,646
  • 4
    Are you sure it is the right integral? It would make much more sense if the denominator was $1-2\alpha\cos(x)\color{red}{+\alpha^2}$. – Jack D'Aurizio Dec 27 '15 at 01:17
  • Please unblock it. Jack D'Aurizio's comment corrects the typo. This integral appears in the Wikipedia article https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign in its example 3. With the typo corrected, the integral can be solved. – Amey Joshi Dec 27 '15 at 03:18

5 Answers5

6

Let us write the integrand as \begin{equation} g(x) = \frac{1 - \alpha^2}{1 - 2\alpha\cos x + \alpha^2} \end{equation} 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 \begin{equation} g(x) = \frac{(1 - \alpha^2)\sec^2(x/2)}{(1 - \alpha)^2 + (1 + \alpha)^2\tan^2(x/2)} \end{equation} which is same as \begin{equation} g(x) = 2\frac{1 + \alpha}{2}\frac{(1 - \alpha)\sec^2(x/2)}{(1 - \alpha)^2 + (1 + \alpha)^2\tan^2(x/2)} \end{equation} or, \begin{equation} 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} \end{equation} or \begin{equation} 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] \end{equation} Therefore, \begin{equation} \int g(x)dx = 2\tan^{-1}\left(\frac{1 + \alpha}{1 - \alpha}\tan\left(\frac{x}{2}\right)\right) + c, \end{equation} where $c$ is a constant of integration. Therefore, \begin{equation} \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} \end{equation}

Amey Joshi
  • 1,084
4

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 $$I = \pi \mathop{\rm Res}_{z=1/\alpha} \frac{1-\alpha^2}{d} = -\pi.$$

Aritra Das
  • 3,528
Fabian
  • 23,360
1

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 $-\pi$.

Community
  • 1
Paramanand Singh
  • 87,309
0

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!

ALIX
  • 1
0

On page 393, Entry 3.613, in the monograph

I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8,

the formula \begin{equation*} \int_0^\pi\frac{\cos(nx)\operatorname{d}x}{1-2a\cos x+a^2}= \begin{cases} \dfrac{\pi a^n}{1-a^2}, & a^2<1\\ \dfrac{\pi}{(a^2-1)a^n}, & a^2>1 \end{cases} \end{equation*} for $n\ge0$ is listed. Consequently, taking $n=0$ gives the result \begin{gather*} \int_0^\pi\frac{(1-a^2)\operatorname{d}x}{1-2a\cos x+a^2}= (1-a^2)\int_0^\pi\frac{\operatorname{d}x}{1-2a\cos x+a^2}\\ =(1-a^2) \begin{cases} \dfrac{\pi}{1-a^2}, & a^2<1\\ \dfrac{\pi}{(a^2-1)}, & a^2>1 \end{cases} =\begin{cases} \pi, & a^2<1\\ -\pi, & a^2>1 \end{cases} \end{gather*} for $a^2\ne1$.

qifeng618
  • 1,691