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I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$: $$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$

which gives (Wolfram Alpha)

$$D= \left[ \frac{\sin \theta(\cos \theta - a)}{2(a^2-1)(a \cos\theta-1)^2} +\frac{\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{(a^2-1)^{3/2}}\right]_0^{2\pi} .$$

which can be expressed as $$D= \left[ \frac{(a^2-1)^{1/2}\sin \theta(\cos \theta - a)+2 (a \cos\theta-1)^2\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{2(a^2-1)^{3/2}(a \cos\theta-1)^2} \right]_0^{2\pi} .$$

This expression involves discontinuities and complex numbers which is beyond my present abilities to handle.

UPDATE

I accepted Jack's answer although I did not follow all the steps. In particular the first equation:-

It is not difficult to check that for any $a\in\mathbb{R}$ such that $|a|<1$ we have $$ \int_{0}^{2\pi}\frac{d\theta}{1-a\cos\theta} = \frac{2\pi}{\sqrt{1-a^2}}\tag{1}$$

but I found an explanation for this in the Kepler-based answer from \u\user5713492 to this question re solving $\int\frac{dx}{a+b\cos{x}}$. Other responders also referred to this equation.

Also interetsing is this article - warning PDF! linked to by \u\Adhvaitha.

steveOw
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  • The equation you got from Wolfram does not seem to be correct. – Jack Feb 08 '18 at 14:26
  • @Bennett oh thanks, my mistake, I have editted it. – steveOw Feb 08 '18 at 15:16
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    The result from Wolfram is $$ \frac{\sqrt{a^2-1} \sin (\theta) (\cos (\theta)-a)+2 (a \cos (\theta)-1)^2 \tanh ^{-1}\left(\frac{(a+1) \tan \left(\frac{\theta}{2}\right)}{\sqrt{a^2-1}}\right)}{2 \left(a^2-1\right)^{3/2} (a \cos (\theta)-1)^2}$$ which is different from yours. – xpaul Feb 08 '18 at 15:39
  • @xpaul yes, sorry I made a mistake using Wolfram Alpha. I have now tidied up the question. – steveOw Feb 08 '18 at 18:39

6 Answers6

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This calls for Kepler's angle! Let $$\sin\theta=\frac{\sqrt{1-a^2}\sin\psi}{1+a\cos\psi}$$ Then $$\begin{align}\cos\theta&=\frac{\cos\psi+a}{1+a\cos\psi}\\ d\theta&=\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\ 1-a\cos\theta&=\frac{1-a^2}{1+a\cos\psi}\end{align}$$ Also when $\theta$ makes a full cycle, so does $\psi$, so $$\begin{align}\int_0^{2\pi}\frac{\sin^2\theta}{\left(1-a\cos\theta\right)^3}d\theta&=\int_0^{2\pi}\frac{\left(1-a^2\right)\sin^2\psi}{\left(1+a\cos\psi\right)^2}\frac{\left(1+a\cos\psi\right)^3}{\left(1-a^2\right)^3}\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\ &=\frac1{\left(1-a^2\right)^{3/2}}\frac12\int_0^{2\pi}\left(1-\cos2\psi\right)d\psi\\ &=\frac1{2\left(1-a^2\right)^{3/2}}\left.\left[\psi-\frac12\sin2\psi\right]\right|_0^{2\pi}=\frac{\pi}{\left(1-a^2\right)^{3/2}}\end{align}$$

user5713492
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Note that $$D = 2\int_0^{\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$$ Let $t=\tan\frac{\theta}2$, we have \begin{align} D&=2\int_0^{\infty} \frac{\frac{4t^2}{(1+t^2)^2}}{(1-a\frac{1-t^2}{1+t^2})^3}\frac{2}{1+t^2}\,dt\\ &=\int_0^{\infty} \frac{16t^2}{(1-a+(1+a)t^2)^3}\,dt\\ &=\frac{\pi}{(1-a^2)^{3/2}}\tag{1} \end{align} $(1):$ for all $a,b>0$, \begin{align} \int_0^{\infty} \frac{16t^2}{(a+bt^2)^3}\,dt&=\frac{1}{(ab)^{3/2}}\int_0^{\infty} \frac{16t^2}{(1+t^2)^3}\,dt\\ &=\frac{1}{(ab)^{3/2}}\int_0^{\infty} {8s^{1/2}}{(1+s)^{-3}}\,ds\tag{$s=t^2$}\\ &=\frac{\pi}{(ab)^{3/2}}\tag{2} \end{align} $(2):$ For $x,y>0$, Euler integral gives $$\int^\infty_0 \frac {t^{x-1}}{(1+t)^{x+y }}dt =\frac {\Gamma (x)\Gamma (y)}{\Gamma (x+y)}$$ and for $x=y=\frac32$, we have $$\frac {\Gamma^2 (\frac32)}{\Gamma (3)}=\frac\pi8$$

Aforest
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It is not difficult to check that for any $a\in\mathbb{R}$ such that $|a|<1$ we have $$ \int_{0}^{2\pi}\frac{d\theta}{1-a\cos\theta} = \frac{2\pi}{\sqrt{1-a^2}}\tag{1}$$ hence by differentiating both sides of $(1)$ with respect to $a$ we have $$ \int_{0}^{2\pi}\frac{\cos(\theta)\,d\theta}{(1-a\cos\theta)^2}=\frac{2a\pi}{(1-a^2)^{3/2}},\qquad \int_{0}^{2\pi}\frac{d\theta}{(1-a\cos\theta)^2}= \frac{2\pi}{(1-a^2)^{3/2}}\tag{2}$$ and by differentiating again $$ \int_{0}^{2\pi}\frac{\sin^2(\theta)\,d\theta}{(1-a\cos\theta)^3}=\frac{\pi}{(1-a^2)^{3/2}} \tag{3}$$ is simple to prove.

Jack D'Aurizio
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  • how do you go from (2.1) to (2.2)? – steveOw Feb 08 '18 at 22:15
  • @steveOw: $(1)+(2.1)\mapsto (2.2)$, since $1=(1-a\cos\theta)+a\cos\theta$. – Jack D'Aurizio Feb 08 '18 at 22:22
  • aha thanks., and (note to self) also multiplying numerators of (2.1) by $a$ to make (2.1.1) and multiplying RHS of (1) by $(1-a^2)$ to make (1.1) before adding (1.1) and (2.1.1) to make (2.2). – steveOw Feb 08 '18 at 22:56
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    I have to admit to not really liking this answer very much, because A) the steps are really hard to follow (I still don't fully get how step (3) works) B) every integral given in the answer is immediate via Kepler's angle, so why not use Kepler's angle in the first place and C) if you were going to use this approach why not say $$\begin{align}\int_0^{2\pi}\frac{\sin^2x}{(1-a\cos x)^3}dx&=\left.\frac{-1}{2a(1-a\cos x)^2}\sin x\right|_0^{2\pi}+\frac1{2a}\int_0^{2\pi}\frac{\cos x}{(1-a\cos x)^2}dx\&=\frac1{2a}\frac d{da}\int_0^{2\pi}\frac{dx}{(1-a\cos x)}\end{align}$$ Seems simpler to me. – user5713492 Feb 09 '18 at 17:31
  • @user5713492: on the other hand, suggesting the use of Kepler's angle or just the observation that (by differentiation) everything is related to the area enclosed by an ellipse is pretty much the same thing. – Jack D'Aurizio Feb 09 '18 at 18:04
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Let $z=e^{i\theta}$ and then $$ \cos\theta=\frac12(z+z^{-1}),\sin\theta=\frac1{2i}(z-z^{-1}),d\theta=\frac{1}{iz}dz. $$ So \begin{eqnarray} &&\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta\\ &=&\int_{|z|=1}\frac{[\frac1{2i}(z-z^{-1})]^2}{(1-\frac a2(z+z^{-1}))^3}\frac{1}{iz}dz\\ &=&-\int_{|z|=1}\frac{2 i \left(z^2-1\right)^2}{\left(a z^2+a-2 z\right)^3}dz\\ &=&-\frac{2i}{a^3}\int_{|z|=1}\frac{\left(z^2-1\right)^2}{\left(z^2-\frac{2}{a} z+1\right)^3}dz\\ &=&-\frac{2i}{a^3}\int_{|z|=1}\frac{\left(z^2-1\right)^2}{\left(z-\frac{1+\sqrt{1-a^2}}{a}\right)^3\left(z-\frac{1-\sqrt{1-a^2}}{a}\right)^3}dz\\ &=&-\frac{2i}{a^3}\cdot2\pi i\cdot\frac12\frac{d^2}{dz^2}\frac{\left(z^2-1\right)^2}{\left(z-\frac{1+\sqrt{1-a^2}}{a}\right)^3}\bigg|_{z=\frac{1-\sqrt{1-a^2}}{a}}\\ &=&\frac{4\pi}{a^3}\cdot\frac{a^3}{4(1-a^2)^{3/2}}\\ &=&\frac{\pi}{(1-a^2)^{3/2}}. \end{eqnarray}

xpaul
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We can use the standard formula $$\int_{0}^{2\pi}\frac{dx}{1-a\cos x} =\frac{2\pi}{\sqrt{1-a^2}},|a|<1\tag{1}$$ Note that $$\frac{d} {dx} \frac{\sin x} {1-a\cos x} =\frac{\cos x-a} {(1-a\cos x) ^2} =-\frac{1}{a}\frac{a^2-1+1-a\cos x} {(1-a\cos x) ^2}$$ and integrating the above with respect to $x$ in interval $[0,2\pi]$ we get $$0=\frac{1-a^2}{a}\int_{0}^{2\pi}\frac{dx}{(1-a\cos x) ^2}-\frac{1}{a}\int_{0}^{2\pi}\frac{dx}{1-a\cos x} $$ ie $$\int_{0}^{2\pi}\frac{dx}{(1-a\cos x) ^2}=\frac{2\pi}{(1-a^2)^{3/2}}\tag{2}$$ Next we can observe that $$\frac{d} {dx} \frac{\sin x} {(1-a\cos x) ^2}=\frac{\cos x-a-a\sin^2x}{(1-a\cos x) ^3}$$ and integrating the above we obtain $$0=-aI+\frac{1-a^2}{a}\cdot J-\frac{1}{a}\cdot\frac{2\pi}{(1-a^2)^{3/2}}$$ ie $$(1-a^2)J-a^{2}I=\frac{2\pi}{(1-a^2)^{3/2}}\tag{3}$$ where $I$ is the integral in question and $J=\int_{0}^{2\pi}(1-a\cos x) ^{-3}\,dx$. We need another relationship between $I, J$ to evaluate both of them. Note that if $t=\cos x$ then $$\frac{\sin^2x}{(1-a\cos x) ^3}=\frac{1-t^2}{(1-at)^3}=\frac{A} {1-at}+\frac{B}{(1-at)^{2}}+\frac{C}{(1-at)^3}\tag{4}$$ where $$A=-\frac{1}{a^2},B=\frac{2}{a^2},C=-\frac{1-a^2}{a^2}$$ and on integrating equation $(4)$ we get $$a^2I+(1-a^2)J=\frac{2\pi(1+a^2)}{(1-a^2)^{3/2}}\tag{5}$$ Subtracting equation $(3)$ from equation $(5)$ we get $$I=\frac{\pi} {(1-a^2)^{3/2}}$$ Comparing this elementary solution with the one offered by Jack D'Aurizio shows us the power of Feynman's technique.

Formula $(1)$ is an easy consequence of the following result $$\int_{0}^{\pi}\frac{dx}{a-b\cos x} =\frac{\pi} {\sqrt{a^2-b^2}},a>|b|\tag{6}$$ which is (not so) easily proved via the substitution $$(a-b\cos x) (a+b\cos y) =a^2-b^2$$ The same substitution $$(1-a\cos x) (1+a\cos y) =1-a^2$$ has been used directly in this answer to evaluate the integral in question with much less effort.

Paramanand Singh
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  • I appreciate the detailed workings. Is there an explanatory source for equation (1) or should I ask another question? – steveOw Sep 29 '23 at 14:40
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    @steveOw: I have given the substitution to prove formula $(1)$ in later part of the answer. Do try that substitution. – Paramanand Singh Sep 29 '23 at 19:10
  • Oh silly me! Thanks, I will will give it a go! Btw shouldn't those two $\cos y$ be $\cos x$? – steveOw Sep 29 '23 at 22:58
  • @steveOw: no, the equations give the relationship between variables $x$ and $y$ using which the integral is changed from $x$ variable to $y$ variable. – Paramanand Singh Sep 30 '23 at 03:16
  • Oh thanks. I think I'm out of my depth here, it is all too cryptic for me, I think I should ask it as another question. – steveOw Sep 30 '23 at 09:42
  • FYI I have added an UPDATE to my question linking to a Keplerian solution in an answer by \u\user5713492 to another question. – steveOw Sep 30 '23 at 16:19
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By integration by parts, we have $$ \begin{aligned}\int_0^{2 \pi} \frac{\sin ^2 x}{(1-a \cos x)^3} d x =- & \frac{1}{2 a} \int_0^{2 \pi} \sin x d\left[\frac{1}{(1-a \cos x)^2}\right] \\ = & {\left[-\frac{\sin x}{2 a(1-a \cos x)^2}\right]_0^{2 \pi}+\frac{1}{2 a} \int_0^{2 \pi} \frac{\cos x}{(1-a \cos x)^2} d x } \\ = & \frac{1}{2 a} \frac{d}{d a}\left[\int_0^{2 \pi} \frac{1}{1-a \cos x} d x\right] \\ = & \frac{1}{2 a} \cdot \frac{d}{d a}\left(\frac{2 \pi}{\sqrt{1-a^2}}\right) \\ = & \frac{1}{2 a} \cdot \frac{2 a \pi}{\left(1-a^2\right)^{\frac{3}{2}}} \\ = & \frac{\pi}{\left(1-a^2\right)^{\frac{3}{2}}} \end{aligned} $$

Lai
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