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I'm dealing with this integral in my spare time, since days and days, and it's really interesting. I'll provide to write what I tried until now, and I would really appreciate some help in understanding how to continue.

$$\Phi(\epsilon) = \frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2}$$

Note: The integral arises in studying the Kepler's problem, which is why $\epsilon\in [0;\ 1]$.

I started with the passages to the complex plane: $$\sin\theta = \frac{1}{2i}(z - z^{-1}) ~~~~~~~ \cos\theta = \frac{1}{2}(z + z^{-1}) ~~~~~~~ \text{d}\theta = \frac{\text{d}z}{iz}$$

thence

$$\begin{align*} \Phi(\epsilon) & = \frac{1}{2\pi}\oint_{|z| = 1}\ \frac{-\frac{1}{4}(z - z^{-1})^2\ \text{d}z}{iz\left[1 + \epsilon\frac{(z + z^{-1}}{2}\right]^2} \\\\ & = -\frac{1}{8\pi i}\oint_{|z| = 1}\ \frac{(z - z^{-1})^2\ \text{d}z}{z\left[\frac{2 + \epsilon(z + z^{-1})}{2}\right]^2} \\\\ & = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{(z - z^{-1})^2\ \text{d}z}{z(2 + \epsilon(z + z^{-1})^2} \\\\ & = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{\left(\frac{z^2 - 1}{z}\right)^2\ \text{d}z}{z\left(2 + \epsilon\frac{z^2+1}{z}\right)^2} \\\\ & = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{(z^2-1)^2\ \text{d}z}{z(2z + \epsilon(z^2+1))^2} \end{align*}$$

Poles et cetera

Poles are at $z_0 = 0$ and when $2z + \epsilon(z^2+1) = 0$, namely

$$z_1 = \frac{1+\sqrt{1 - \epsilon^2}}{\epsilon} ~~~~~~~~~~~ z_2 = \frac{1-\sqrt{1 - \epsilon^2}}{\epsilon}$$

Checking that $z_1\cdot z_2 = 1$ is ok, the question is now:

Question 1.: which one, aside $z_0$ lies into the unitary circle?

Question 2.: How to proceed?

user5389726598465
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Enrico M.
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1 Answers1

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I am assuming $0 < \epsilon < 1$ in the following. The cases $\epsilon = 0$ and $\epsilon = 1$ have to be handled separately, see below.

There is a small error in your calculation, the root of $2z + \epsilon(z^2+1) = 0$ are $$ z_1 = \frac{-1-\sqrt{1 - \epsilon^2}}{\epsilon} \, , \quad z_2 = \frac{-1+\sqrt{1 - \epsilon^2}}{\epsilon} $$

Question 1.: which one, aside $z_0$ lies into the unitary circle?

From $$ 1+\sqrt{1 - \epsilon^2} > 1 > \epsilon $$ it follows that $z_1 < -1$ and consequently, $-1 < z_2 < 0$, i.e. $z_2$ is inside the unit disk and $z_1$ outside.

Question 2.: How to proceed?

You already have that $$ \Phi(\epsilon) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z^2-1)^2\ }{z(2z + \epsilon(z^2+1))^2} \, dz = \frac{1}{2\pi i} \oint_{|z| = 1} f(z) \, dz $$ with $$ f(z) := -\frac{1}{2 \epsilon^2} \frac{(z^2-1)^2\ }{z(z-z_1)^2(z-z_2)^2} $$ From the residue theorem it follows that $$ \Phi(\epsilon) = \text{Res}(f, 0) + \text{Res}(f, z_2) \, . $$ $f$ has a simple pole at $z = 0$, therefore $$ \text{Res}(f, 0) = \lim_{z \to 0} z f(z) = -\frac{1}{2 \epsilon^2} \, . $$ $f$ has a double pole at $z = z_2$. One possible method to compute the residue is the limit formula for higher order poles: $$ \text{Res}(f, z_2) = \lim_{z \to z_2} \frac{d}{dz} \bigl((z-z_2)^2 f(z) \bigr) \, . $$

Special cases: For $\epsilon = 0$ the integral becomes $$ \Phi(0) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z^2-1)^2}{2z^3} \, dz = -\frac{1}{8\pi i}\oint_{|z| = 1} \bigl( z - \frac 2z + \frac{1}{z^3}\bigr) \, dz $$ which can easily be computed using the residue theorem. Alternatively, $$ \Phi(0) = \frac{1}{2\pi}\int_0^{2\pi} \sin^2\theta \, d\theta $$ can be computed directly.

For $\epsilon = 1$, the integral becomes $$ \Phi(1) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z-1)^2}{z(z+1)^2} \, dz $$ which is infinite due to the singularity at $z=-1$. Alternatively, $$ \Phi(1) = \frac{1}{2\pi}\int_0^{2\pi} \frac{\sin^2\theta}{(1 + \cos\theta)^2} \, d\theta = \frac{1}{2\pi}\int_0^{2\pi} \frac{(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})^2 }{(2 \cos^2 \frac{\theta}{2})^2} \, d\theta = \frac{1}{2\pi}\int_0^{2\pi} \tan^2 \frac{\theta}{2}\, d\theta $$ is infinite.

Martin R
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