I'm having a hard time trying to resolve this integral : $$\int_{0}^{2\pi}\frac{\mathrm dx}{(1-a\cos x)^2}$$ where $a$ is a positive real constant.

I tried using substitution, but I'm stuck by the fact that the integral must be computed between $0$ and $2\pi$, which leads to integration between $X$ and $X$ (where $X$ can be $0$ or $1$ anything following the substitution in the boundaries). I'm not looking for the full result if the computation is complicated, just some lead to start off...


UPDATE : seeing the answers proposed, I checked that indeed $|a|<1$.

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    $\begingroup$ Do you realize it essentially is the area enclosed by an ellipse? $\endgroup$ – Jack D'Aurizio Jan 16 at 17:24
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    $\begingroup$ Similar : math.stackexchange.com/q/1846774/321264. $\endgroup$ – StubbornAtom Jan 16 at 18:01
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    $\begingroup$ You can start with the standard result $$\int_{0}^{2\pi}\frac{dx}{a-b\cos x} =\frac{2\pi}{\sqrt{a^2-b^2}}$$ and differentiate it with respect to $a$ and then put $a=1$ and replace $b$ by $a$. $\endgroup$ – Paramanand Singh Jan 17 at 5:24

The polar equation of an ellipse (with respect to a focus, taking the aphelion as the point associated to $\theta=0$) is $$ \rho(\theta) = \frac{\ell}{1-e\cos\theta} $$ where, in terms of the standard parameters, $\ell$ is the semi-latus rectum $\frac{b^2}{a}$ and $e$ is the eccentricity $\frac{c}{a}$.
The area enclosed by such ellipse is $\pi a b $ or $$ \frac{1}{2}\int_{0}^{2\pi}\frac{\ell^2}{(1-e\cos\theta)^2}\,d\theta, $$ so we get that for any $e\in[0,1)$ the following identity holds: $$ \int_{0}^{2\pi}\frac{d\theta}{(1-e\cos\theta)^2}= \frac{2\pi a b}{\ell^2}=2\pi\left(\frac{a}{b}\right)^3=\frac{2\pi}{\left(\frac{b^2}{a^2}\right)^{3/2}}={\frac{2\pi}{(1-e^2)^{3/2}}}. $$ The same holds if, in the LHS, we replace $e$ with $-e$, since $ \rho(\theta) = \frac{\ell}{1+e\cos\theta} $ is the polar equation of the same ellipse, with the perihelion being taken as the point associated to $\theta=0$. So we get that for any $a\in(-1,1)$ $$ \int_{0}^{2\pi}\frac{d\theta}{(1-a\cos\theta)^2}= \color{blue}{\frac{2\pi}{(1-a^2)^{3/2}}}. $$

  • $\begingroup$ Thanks a lot for your answer, I did not realized this was the polar equation of an ellipse! This solution looks quite elegant. I also update my post, the coefficient $a$ is indeed always less than 1. $\endgroup$ – mocquin Jan 16 at 20:33
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    $\begingroup$ +1 for polar equation of ellipse. Although I believe computation of area of ellipse is easier if one uses rectangular coordinates. So in effect your answer uses the equivalence/transformation between these coordinate systems. $\endgroup$ – Paramanand Singh Jan 17 at 5:15


Note that for $|a|<1$

$$\begin{align} \int_0^{2\pi}\frac{1}{(1-a\cos(\theta))^2}\,d\theta&=\frac1{a^2}\int_0^{2\pi}\frac{1}{(1/a-\cos(\theta))^2}\,d\theta\\\\ &=-\frac1{a^2}\frac{d}{d(1/a)}\int_0^{2\pi}\frac{1}{1/a-\cos(\theta)}\,d\theta\tag1\\\\ &= -\frac2{a^2}\frac{d}{d(1/a)}\int_0^{\pi}\frac{1}{1/a-\cos(\theta)}\,d\theta\tag2 \end{align}$$

Now use the Weierstrass substitution to evaluate the integral on the right-hand side of $(2)$ and differentiate to obtain the coveted result.

Alternatively, one could use contour integration to evaluate the integral of interest or evaluate the integral on the right-hand side of $(1)$.

  • $\begingroup$ The Weierstrass substitution is going to fail if you apply it to this integral. You have to do something first (integrate on $[-\pi,\pi]$ and split, for instance). $\endgroup$ – Jean-Claude Arbaut Jan 16 at 17:51
  • $\begingroup$ @Jean-ClaudeArbaut This was meant to be a HINT only. But I've added another line to facilitate. $\endgroup$ – Mark Viola Jan 16 at 19:52

My goal with this answer is to give you a bunch of different really general integral formulas that you may find very useful.

$$F(a)=\int_0^{\pi}\frac{\mathrm dx}{(1+a\cos x)^2}$$ So your integral is $2F(-a)$. First off we will preform the tangent-half-angle substitution $t=\tan(x/2)$. Hence we have that $$F(a)=2\int_0^\infty \frac1{(1+a\frac{1-t^2}{1+t^2})^2}\frac{\mathrm dt}{1+t^2}$$ $$F(a)=2\int_0^{\infty}\frac{1+t^2}{[(1-a)t^2+a+1]^2}\mathrm dt$$ Which we can rewrite as $$F(a)=\frac2{1-a}\int_0^\infty\frac{\mathrm dt}{(1-a)t^2+a+1}+\frac{4a}{a-1}\int_0^{\infty}\frac{\mathrm dt}{[(1-a)t^2+a+1]^2}$$

Next we consider the very general integral $$I(m;a,b)=\int_0^\infty \frac{\mathrm dx}{(ax^2+b)^{m+1}}$$ First we integrate by parts with $\mathrm dv=\mathrm dx$ to produce $$I(m;a,b)=\frac{x}{(ax^2+b)^{m+1}}\bigg|_0^{\infty}+2(m+1)\int_0^\infty\frac{ax^2}{(ax^2+b)^{m+2}}\mathrm dx$$ $$I(m;a,b)=2(m+1)\int_0^\infty\frac{ax^2+b-b}{(ax^2+b)^{m+2}}\mathrm dx$$ $$I(m;a,b)=2(m+1)I(m;a,b)-2(m+1)bI(m+1;a,b)$$ Then solving for $I(m+1;a,b)$ and replacing $m+1$ with $m$, $$I(m;a,b)=\frac{2m-1}{2bm}I(m-1;a,b)$$ And for the base case: $$I(0;a,b)=\int_0^{\infty}\frac{\mathrm dx}{ax^2+b}$$ The Trig sub $x=\sqrt{\frac{b}a}\tan u$ gives $$I(0;a,b)=\frac\pi{2\sqrt{ab}}$$ Which is a special case of $$\int_{x_1}^{x_2}\frac{\mathrm dx}{ax^2+bx+c}=\frac2{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}\bigg|_{x_1}^{x_2}$$ Anyway, the recurrence has the solution $$I(m;a,b)=\frac\pi{2^{2m+1}b^m\sqrt{ab}}{2m\choose m}$$ Which we will apply very shortly!

Recalling the definition of $I(m;a,b)$, we have that $$F(a)=\frac2{1-a}I(0;1-a,1+a)+\frac{4a}{a-1}I(1;1-a,1+a)$$ $$F(a)=\frac\pi{(1-a^2)^{3/2}}$$ And since your integral is given by $2F(-a)$, $$\int_0^{2\pi}\frac{\mathrm dx}{(1-a\cos x)^2}=\frac{2\pi}{(1-a^2)^{3/2}}$$


Consider the integral $$C(n;a)=\int_0^\pi \frac{\mathrm dx}{(1+a\cos x)^n}$$ Starting with $t=\tan\frac{x}2$, $$C(n;a)=2\int_0^\infty \frac1{\left[1+a\frac{1-t^2}{1+t^2}\right]^n}\frac{\mathrm dt}{1+t^2}$$ $$C(n;a)=2\int_0^\infty \frac{\left(t^2+1\right)^{n-1}}{\left[(1-a)t^2+1+a\right]^n}\mathrm dt$$ Then using the binomial theorem, $$C(n;a)=2\sum_{k=0}^{n-1}{n-1\choose k}\int_0^\infty\frac{x^{2k}}{\left[(1-a)x^2+1+a\right]^n}\mathrm dx$$ $$C(n;a)=\frac2{(1-a)^n}\sum_{k=0}^{n-1}{n-1\choose k}\int_0^\infty\frac{x^{2k}}{\left[x^2+\frac{1+a}{1-a}\right]^n}\mathrm dx$$ We then recall the integral due to my collaborator @DavidG, $$\int_0^\infty\frac{x^{q}}{\left[x^w+b\right]^p}\mathrm dx=\frac{b^{\frac{1+q}{w}-p}}{w}\frac{\Gamma\left(p-\frac{1+q}{w}\right)\Gamma\left(\frac{1+q}{w}\right)}{\Gamma\left(p\right)}$$ Here $\Gamma(s)$ is the Gamma function. So with $w=2$, $b=\frac{1+a}{1-a}$, $p=n$, and $q=2k$, $$C(n;a)=\frac1{(1-a)^n}\sum_{k=0}^{n-1}{n-1\choose k}\left(\frac{1+a}{1-a}\right)^{\frac{2k+1}{2}-n}\frac{\Gamma\left(n-\frac{2k+1}{2}\right)\Gamma\left(\frac{2k+1}{2}\right)}{\Gamma\left(n\right)}$$ $$C(n;a)=\frac1{(1-a)^n}\sum_{k=0}^{n-1}\left(\frac{1+a}{1-a}\right)^{\frac{2k+1}{2}-n}\frac{\Gamma\left(n-\frac{2k+1}{2}\right)\Gamma\left(\frac{2k+1}{2}\right)}{k!\Gamma(n-k)}$$


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