5
$\begingroup$

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.
$\endgroup$
  • 4
    $\begingroup$ Are you sure it is the right integral? It would make much more sense if the denominator was $1-2\alpha\cos(x)\color{red}{+\alpha^2}$. $\endgroup$ – Jack D'Aurizio Dec 27 '15 at 1:17
  • $\begingroup$ 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. $\endgroup$ – Amey Joshi Dec 27 '15 at 3:18
6
$\begingroup$

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}

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Residue theorem (almost always) gives an elegant solution. Thanks. $\endgroup$ – Amey Joshi Dec 27 '15 at 7:31
1
$\begingroup$

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 $-\pi$.

$\endgroup$
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.