Integral challenge ∫(sin^2θ)/(1−2acosθ+a^2)dθ, 0<a<1

[lfdahl]

Solve the definite integral

\[I(a) = \int_{0}^{2\pi}\frac{\sin^2 \theta }{1-2a\cos \theta + a^2}\: \: d\theta,\;\;\; 0<a<1.\]

 

[DrWahoo]

Solve the definite integral

\[I(a) = \int_{0}^{2\pi}\frac{\sin^2 \theta }{1-2a\cos \theta + a^2}\: \: d\theta,\;\;\; 0<a<1.\]

Nice problem. I will give other members a chance to solve before I do so.

Spoiler

However, a useful hint for others may be the following identity.

$\sin^2(x) = (1 − \cos(2x))/2$
The rest should follow from integral techniques.

If no one answers by tomorrow, I will post up a written solution.

 

[Opalg]

Solve the definite integral

\[I(a) = \int_{0}^{2\pi}\frac{\sin^2 \theta }{1-2a\cos \theta + a^2}\: \: d\theta,\;\;\; 0<a<1.\]

[sp]Solution using complex analysis:

Let $z = e^{i\theta}.$ Then $\sin\theta = \frac1{2i}(z - z^{-1})$, $1 - 2a\cos\theta + a^2 = (z-a)(z^{-1}-a)$ and $dz = iz\,d\theta$. So $$I(a) = \oint_\Gamma \frac{(z - z^{-1})^2}{-4(z-a)(z^{-1}-a)}\frac{dz}{iz} = \frac i4\oint_\Gamma \frac{(z^2-1)^2}{z^2(z-a)(1-za)}\,dz,$$ where $\Gamma$ denotes the unit circle. The meromorphic function \(\displaystyle \frac{(z^2-1)^2}{z^2(z-a)(1-za)}\) has two poles inside $\Gamma$, a double pole at $z=0$ and a simple pole at $z=a$. The residue at $z=0$ is given by $$\def\res{\operatorname{res}}\res(0) = \lim_{z\to0}\frac d{dz}\frac{(z^2-1)^2}{(z-a)(1-za)} = \lim_{z\to0}\frac{4z(z^2-1)(z-a)(1-za) - (z^2-1)^2(1+a^2 - 2az)}{(z-a)^2(1-za)^2} = \frac{-1-a^2}{a^2}.$$ The residue at $z=a$ is given by $$\res(a) = \lim_{z\to a}\frac{(z^2-1)^2}{z^2(1-az)} = \frac{(1-a^2)^2}{a^2(1-a^2)} = \frac{1-a^2}{a^2}.$$ By Cauchy's residue theorem, $$I(a) = \frac i4 2\pi i\bigl(\res(0) + \res(a)\bigr) = -\frac\pi2\Bigl(\frac{-1-a^2)}{a^2} + \frac{1-a^2}{a^2}\Bigr) = \pi.$$

There must be a deeper reason why the result is independent of $a$, presumably something to do with the Poisson kernel?
[/sp]

 

[lfdahl]

[sp]Solution using complex analysis:

Let $z = e^{i\theta}.$ Then $\sin\theta = \frac1{2i}(z - z^{-1})$, $1 - 2a\cos\theta + a^2 = (z-a)(z^{-1}-a)$ and $dz = iz\,d\theta$. So $$I(a) = \oint_\Gamma \frac{(z - z^{-1})^2}{-4(z-a)(z^{-1}-a)}\frac{dz}{iz} = \frac i4\oint_\Gamma \frac{(z^2-1)^2}{z^2(z-a)(1-za)}\,dz,$$ where $\Gamma$ denotes the unit circle. The meromorphic function \(\displaystyle \frac{(z^2-1)^2}{z^2(z-a)(1-za)}\) has two poles inside $\Gamma$, a double pole at $z=0$ and a simple pole at $z=a$. The residue at $z=0$ is given by $$\def\res{\operatorname{res}}\res(0) = \lim_{z\to0}\frac d{dz}\frac{(z^2-1)^2}{(z-a)(1-za)} = \lim_{z\to0}\frac{4z(z^2-1)(z-a)(1-za) - (z^2-1)^2(1+a^2 - 2az)}{(z-a)^2(1-za)^2} = \frac{-1-a^2}{a^2}.$$ The residue at $z=a$ is given by $$\res(a) = \lim_{z\to a}\frac{(z^2-1)^2}{z^2(1-az)} = \frac{(1-a^2)^2}{a^2(1-a^2)} = \frac{1-a^2}{a^2}.$$ By Cauchy's residue theorem, $$I(a) = \frac i4 2\pi i\bigl(\res(0) + \res(a)\bigr) = -\frac\pi2\Bigl(\frac{-1-a^2)}{a^2} + \frac{1-a^2}{a^2}\Bigr) = \pi.$$

There must be a deeper reason why the result is independent of $a$, presumably something to do with the Poisson kernel?
[/sp]

Thankyou very much, Opalg!, for such a thourough, in depth analysis and solution to the problem.
I am always grateful for your participation and highly valuable contributions to this forum.

 

[lfdahl]

An alternative solution:

Spoiler

Let the point $A(a,0)$ be on the $x-axis$ and $P(\cos \theta,\sin \theta)$ on the unit circle.
Let $Q$ be the perpendicular projection of $P$ onto the $x-axis$. Let $\angle PAQ$ be denoted by $\varphi$.
We have, that
\[\frac{\sin ^2\theta }{1-2a \cos \theta + a^2}= \frac{\sin ^2\theta }{\sin ^2\theta +(a-\cos \theta)^2} = \frac{PQ^2}{PQ^2+AQ^2}=\frac{PQ^2}{AP^2}\\\\=\sin^2\varphi=\frac{1-\cos 2\varphi}{2}\]

Hence
\[\int_{0}^{2\pi }\frac{\sin ^2\theta }{1-2a \cos \theta + a^2}\: \: d\theta = \int_{0}^{2\pi }\frac{1-\cos 2\varphi}{2}\: \: d\varphi = \pi.\]

 

[Opalg]

An alternative solution:

Spoiler

Let the point $A(a,0)$ be on the $x-axis$ and $P(\cos \theta,\sin \theta)$ on the unit circle.
Let $Q$ be the perpendicular projection of $P$ onto the $x-axis$. Let $\angle PAQ$ be denoted by $\varphi$.
We have, that
\[\frac{\sin ^2\theta }{1-2a \cos \theta + a^2}= \frac{\sin ^2\theta }{\sin ^2\theta +(a-\cos \theta)^2} = \frac{PQ^2}{PQ^2+AQ^2}=\frac{PQ^2}{AP^2}\\\\=\sin^2\varphi=\frac{1-\cos 2\varphi}{2}\]

Hence
\[\int_{0}^{2\pi }\frac{\sin ^2\theta }{1-2a \cos \theta + a^2}\: \: d\theta = \int_{0}^{2\pi }\frac{1-\cos 2\varphi}{2}\: \: d\varphi = \pi.\]

[sp]That looks like a very neat solution. But I am not understanding what happens when you make the substitution from $\theta$ to $\varphi$ in the integral. Firstly, I think that the limits of integration should change: as $\theta$ goes from $0$ to $2\pi$, it looks as though $\varphi$ should go from $\pi$ to $-\pi$. Secondly, why should $d\varphi$ be the same as $d\theta$? The relationship between $\theta$ and $\varphi$ seems quite complicated to me.

I am encouraged by the fact that the answer comes out the same as I found, but I wonder whether some further explanation is needed?

[/sp]

 

[lfdahl]

[sp]That looks like a very neat solution. But I am not understanding what happens when you make the substitution from $\theta$ to $\varphi$ in the integral. Firstly, I think that the limits of integration should change: as $\theta$ goes from $0$ to $2\pi$, it looks as though $\varphi$ should go from $\pi$ to $-\pi$. Secondly, why should $d\varphi$ be the same as $d\theta$? The relationship between $\theta$ and $\varphi$ seems quite complicated to me.

I am encouraged by the fact that the answer comes out the same as I found, but I wonder whether some further explanation is needed?

[/sp]

Spoiler

Hi, Opalg! Thankyou for your considerations. I must admit, I was neither aware of the possible change in the limits nor the hidden possibly complicated relationship between $d\theta$ and $d\varphi$.
As for the limits, it seems that $\varphi$ is confined to the interval $[0;\frac{\pi}{2}]$ starting from $0$ radians four times: two, when $Q$ coincides with $A$, and two when $P$ is located on the $x$-axis. Am I right?
Maybe you could help reveal the relation between $d\theta$ and $d\varphi$?
I am sorry, that I did not accommodate these issues from the start, when I posted the alternative solution.
I was much encouraged by the fact that the alternative solution had the same answer as your solution.
But again: that doesn´t justify the straight forward change of the variables.

 

[DrWahoo]

Spoiler

Hi, Opalg! Thankyou for your considerations. I must admit, I was neither aware of the possible change in the limits nor the hidden possibly complicated relationship between $d\theta$ and $d\varphi$.
As for the limits, it seems that $\varphi$ is confined to the interval $[0;\frac{\pi}{2}]$ starting from $0$ radians four times: two, when $Q$ coincides with $A$, and two when $P$ is located on the $x$-axis. Am I right?
Maybe you could help reveal the relation between $d\theta$ and $d\varphi$?
I am sorry, that I did not accommodate these issues from the start, when I posted the alternative solution.
I was much encouraged by the fact that the alternative solution had the same answer as your solution.
But again: that doesn´t justify the straight forward change of the variables.

Rossenwasser has a nice book on this type of interchanging order differentiation/integration. I think the title is dynamic modeling of autonomous control systems. He uses this technique to interchange the order of differentiation, under the assumption the function at hand is continuous and differentiable at said point.

He really doesn't go into much of the alg behind it, but he provides a nice proof in his book, which I am too lazy to type out. Great book for modeling complex systems.

 

[lfdahl]

Rossenwasser has a nice book on this type of interchanging order differentiation/integration. I think the title is dynamic modeling of autonomous control systems. He uses this technique to interchange the order of differentiation, under the assumption the function at hand is continuous and differentiable at said point.

He really doesn't go into much of the alg behind it, but he provides a nice proof in his book, which I am too lazy to type out. Great book for modeling complex systems.

Thankyou, DrWahoo. for your good advice!
By the way: You wrote, that you also have a solution to the challenge. Maybe, you will share it in the forum?

 

[lfdahl]

Thanks to Opalg the alternative solution above (please cf. #5) has been analyzed in more detail with very interesting comments from him, and I am happy to share them with the followers in this thread:

Spoiler

I have tried to evaluate that integral by the method you [lfdahl] used, but I have been totally unable to make it work. The angles $\theta$ and $\varphi$ are two of the angles of the triangle $OPA$, but there does not seem to be a usable formula to connect them.

Wolfram Alpha gives the indefinite integral as $$\int \frac{\sin^2\theta}{1 - 2a\cos\theta + a^2}\,d\theta = \frac1{4a^2}\Bigl(\theta + a^2\theta + 2(1-a^2)\tan^{-1}\bigl(\tfrac{a+1}{a-1}\tan\tfrac \theta2\bigr) + 2a\sin\theta \Bigr) + \text{const.}$$ I have not tried to check that manually, but presumably it comes from the substitution $t = \tan\frac\theta2.$ That formula may not be safe for evaluating the definite integral from $0$ to $2\pi$ because $\tan\frac\theta2$ becomes infinite at $\theta = \pi$. But the integral from $\pi$ to $2\pi$ is the same as the integral from $0$ to $\pi$. You can use the Wolfram formula for the integral from $0$ to $\pi$ (by taking the limit as $\theta \to \pi$), and it gives $$\int_0^\pi \frac{\sin^2\theta}{1 - 2a\cos\theta + a^2}\,d\theta = \frac\pi2.$$ This confirms that $I(a)$ is independent of $a$ and is equal to $\pi$.

The advantage of the complex variable method that I used is that it gives the definite integral without having to compute the indefinite integral.