Show that $$\begin{align*} \forall x \in [-1,1]: \int_0^{\pi} \frac{\sin^n \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} \, d\theta &= c_n \tag{1} \\ \int_0^{\pi} \frac{\sin^{n+2} \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} \, d\theta &= a_n \cdot x^2+b_n \tag{2} \end{align*}$$ where $a_n, b_n, c_n$ are constants (which do not depend on $x \in [-1,1]$), $n \geq 3$.

I tried several approaches (differentiation to show that the derivative of $(1)$ is equal to $0$, Weierstraß substitution, ...), but always got stuck. For example, one can show that the integrand in $(1)$ equals

$$\begin{align*} \frac{\sin^n \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} &= \left( \sqrt{ \left( \frac{x-\cos \theta}{\sin \theta} \right)^2+1} \right)^{-n} \\ &= \left( \sqrt{4 \left( \frac{\sin \frac{\theta+\varrho}{2} \cdot \sin \frac{\theta-\varrho}{2}}{\sin \theta} \right)^2+1} \right)^{-n} \end{align*}$$ where $x=\cos \varrho$. I hoped to get some kind of symmetrization out of it, but (as far as I can see) it doesn't work.

Any ideas? (The aim is to find a rather quick or direct proof - a lengthy one is already known, using a recursive/inductive approach.)


  • $\begingroup$ Would the downvoter care to comment? Thanks. $\endgroup$
    – saz
    Mar 10, 2014 at 18:54

2 Answers 2


The Gegenbauer polynomials were built to solve problems like this. Suppose $m\in\mathbb{N}$. We have $$\begin{eqnarray*} \int_0^{\pi} d\theta \, \frac{\sin^{n+2m} \theta}{(1-2x \cos \theta+x^2)^{n/2}} &=& \int_0^\pi d\theta \, \sin^{n+2m} \theta \sum_{k=0}^\infty C_k^{(n/2)}(\cos\theta) x^k \\ &=& \int_{-1}^1 d u \, (1-u^2)^{n/2-1/2} (1-u^2)^m \sum_{k=0}^\infty C_k^{(n/2)}(u) x^k \hspace{5ex} (u=\cos\theta) \\ &=& \int_{-1}^1 d u \, (1-u^2)^{n/2-1/2} \sum_{l=0}^m \beta_{2l} C_{2l}^{(n/2)}(u) \sum_{k=0}^\infty C_k^{(n/2)}(u) x^k \\ &=& \sum_{l=0}^m \sum_{k=0}^\infty \beta_{2l} x^k \underbrace{\int_{-1}^1 d u \, (1-u^2)^{n/2-1/2} C_{2l}^{(n/2)}(u) C_k^{(n/2)}(u)}_{\gamma_{2l}^{(n/2)} \delta_{2l,k}} \\ &=& \sum_{l=0}^m \beta_{2l} \gamma_{2l}^{(n/2)} x^{2l}. \end{eqnarray*}$$ This proves the claim for any $m\in\mathbb{N}$. The restriction $n\ge 3$ is too loose. The result holds for any $n$ such that $n>-1$.

For a given $m$ it is a straightforward exercise to find the $\beta$s and $\gamma$s. For example, using the fact that $1=C_0^{(n/2)}(u)$ and the normalization relation for the Gegenbauer polynomials we find $$\int_0^{\pi} d\theta \, \frac{\sin^{n} \theta}{(1-2x \cos \theta+x^2)^{n/2}} = \frac{\sqrt{\pi}\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}.$$

Some details:

  • $1/(1-2x \cos \theta+x^2)^{n/2}$ is the generating function for the Gegenbauer polynomials.
  • The polynomials have the parity property.
  • The polynomials provide an orthogonal basis for functions on $[-1,1]$ with weight function $(1-u^2)^{n/2-1/2}$.
  • $\begingroup$ That's great, thanks! $\endgroup$
    – saz
    May 18, 2013 at 19:49
  • $\begingroup$ @saz: Glad to help. Thanks for the interesting question. $\endgroup$
    – user26872
    May 18, 2013 at 20:00

You are in the good way. The integral can be expressed as $$ I(\varphi)=\int_{0}^{\pi}f(\theta,\varphi)d\theta $$ where $$ f(\theta,\varphi)=\left[1+\left(\dfrac{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}{\sin\theta}\right)^{2}\right]^{-\frac{n}{2}} $$ where $x=\cos\varphi$.

Now, $f(\theta,\varphi)=f(-\theta,\varphi)$, so is no harm in writing the integral as $$ I(\varphi)=\dfrac{1}{2}\int_{-\pi}^{\pi}f(\theta,\varphi)d\theta $$ We want to prove that $$ \cfrac[l]{\partial I(\varphi)}{\partial\varphi}=\dfrac{1}{2}\int_{-\pi}^{\pi}\cfrac[l]{\partial f(\theta,\varphi)}{\partial\varphi}d\theta=0 $$ so it is enough to prove that $$ g(\theta,\varphi)=\cfrac[l]{\partial f(\theta,\varphi)}{\partial\varphi} $$ is antisymmetrical in $\theta$.

Deriving we obtaing $$ g(\theta,\varphi)=\dfrac{n\left(\dfrac{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}{\sin\theta}\right)}{\left[1+\left(\dfrac{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}{\sin\theta}\right)^{2}\right]^{\frac{n}{2}+1}} $$ and we see that $g(\theta,\varphi)=-g(-\theta,\varphi)$.

For expression (2), the proof is analogous.

  • $\begingroup$ How do you arrive at this expression for $g$? As far as I can see, we have $$g(\theta,\varphi) = \frac{n}{(\ldots)^{\frac{n}{2}+1}} \cdot \frac{2}{\sin \theta} \cdot \frac{\partial}{\partial \varphi} \left(\sin \frac{\theta+\varphi}{2} \cdot \sin \frac{\theta-\varphi}{2} \right)$$ ... along my calculations, I end up with $g(\theta,\varphi) = g(-\theta,\varphi)$. $\endgroup$
    – saz
    May 18, 2013 at 19:40
  • $\begingroup$ Excuse me, Saz. There was a mistake in my calculations. Forget my post. $\endgroup$
    – CarlesV
    May 18, 2013 at 20:00

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