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I am trying to prove that for $0<=a<1$ $$\int_0^{2\pi}{\frac {\sin^2\theta (cos\theta - a)}{(1-a \,cos \theta)^4}}d \theta = 0$$

I know that $$\int_0^{2\pi}{\frac {\sin\theta \,cos\theta}{(1-a \,cos \theta)^4}}d \theta = 0$$

and $$\int_0^{2\pi}{\frac {3a\sin\theta}{(1-a \,cos \theta)^4}}d \theta = \int_0^{2\pi}{\frac {cos\theta}{(1-a \,cos \theta)^3}}d \theta$$

I have tried substituting this last equation into the first but it doesn't seem to produce any obvious simplifications.

steveOw
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2 Answers2

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You can apply a neat complex analysis trick to integrals of the form $$ \int_0^{2\pi} F(\cos \theta, \sin \theta)\ \mathrm{d}\theta. $$ Make the substitution $z = e^{i\theta} = \cos \theta + i \sin \theta$. This transforms the contour of integration from the interval $[0,2\pi]$ into the unit circle $C_1(0)$ lying in the complex plane. We can obtain $\mathrm{d}\theta$ as $$ \mathrm{d}z = i e^{i\theta} \mathrm{d}\theta \implies \mathrm{d}\theta = \frac{\mathrm{d}z}{iz}. $$ Furthermore, observe that $\frac{1}{z} = e^{-i\theta} = \cos \theta - i \sin \theta$, so we can obtain $\cos \theta$ and $\sin \theta$ as $$ \cos \theta = \frac{1}{2} \left( z + \frac{1}{z} \right) \hspace{1cm} \sin \theta = \frac{1}{2i} \left( z - \frac{1}{z} \right). $$ For your integral, this substitution produces $$ \int_0^{2\pi} \frac{\sin^2 \theta (\cos \theta - a)}{(1 - a \cos \theta)^4} \mathrm{d}\theta = \int_{C_1(0)} \frac{2 i \left(z^2-1\right)^2 \left(z^2 - 2az +1\right)}{\left(a z^2 - 2z +a\right)^4} \mathrm{d}z.$$ Can you proceed from here?

David Zhang
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  • .Thanks, I will try...but it make take me some time as I am not familiar with complex integrals. – steveOw Nov 05 '14 at 00:03
  • @steveOw The next step is to apply the Residue Theorem. We want to find the poles of that rational function which lie within the unit circle. There is only one such pole, of order 4, given by $\frac{1-\sqrt{1-a^2}}{a}$. – David Zhang Nov 05 '14 at 00:43
  • I looked at the Residue Theorem but it is too advanced for me personally to tackle at present. – steveOw Nov 05 '14 at 10:21
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The indefinite integral has an elementary closed form. Let $$f_a(\theta) = \frac{\sin^2 \theta (\cos \theta - a)}{(1 - a \cos \theta)^4},$$ and $$u = \frac{1 - a \cos \theta}{\sin \theta}, \quad du = \frac{a - \cos\theta}{\sin^2 \theta} \, d\theta,$$ and thus $$\int f_a(\theta) \, d\theta = -\int u^{-4} \, du = \frac{1}{3} u^{-3} + C = \frac{\sin^3 \theta}{3(1-a \cos\theta)^3} + C.$$ Evaluating this for $\theta = 0$ and $\theta = 2\pi$ shows that the definite integral is zero.

steveOw
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heropup
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  • Thanks that looks nice and simple. I wonder why WolframAlpha (see the end of my question) gives a different indefinite integral formula from yours? – steveOw Nov 05 '14 at 00:32
  • Aha, my error. My expansion (and thus my indefinite integral therefrom) were wrong. – steveOw Nov 05 '14 at 10:18
  • I think your result should have cos in the denominator instead of sin. – steveOw Nov 05 '14 at 11:51