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So I've read a book and found myself stumped in this integral:

$$\int_{0}^{\pi} \frac{\cos(n\theta)}{b^2-a^2\cos(2\theta)}\, d\theta=\begin{cases} \,\,0 &,\quad\mbox{if}\,\, n\,\,\mbox{is odd}\\[20pt] \,\,\dfrac{\pi}{\sqrt{b^4-a^4}}\left(\dfrac{\sqrt{b^2-\sqrt{b^4-a^4}}}{a}\right)^n&,\quad\mbox{if}\,\, n\,\,\mbox{is even}\\ \end{cases}$$ where $b>a$.

Anyone knows how to evaluate it? Or knows a reference for helping me to prove formula above?

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A hint would suffice.

Use the following identity $$ \frac{p^2-q^2}{p^2-2pq\cos x+q^2}=1+2\sum_{n=1}^\infty \left(\frac{q}{p}\right)^n\cos(n x) $$ Determine $p$ and $q$ in term of $a$ and $b$, then use identity $$\int_0^\pi\cos(n x)\cos(mx)\,dx=\begin{cases} \,\,0 &,\quad\mbox{if}\,\, n\ne m\\[12pt] \,\,\dfrac{\pi}{2}&,\quad\mbox{if}\,\, n=m\,\,\mbox{and}\,\,n\neq0\\ \end{cases}$$ You may refer to here for the proof and the reference to help you evaluate the integrals.

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  • $\begingroup$ Do you happen to know the reference text where this calculation is done explicitly? $\endgroup$ – MathematicalPhysicist Nov 1 '14 at 15:16
  • $\begingroup$ @MathematicalPhysicist See my answer for an explicit and IMO more standard calculation. $\endgroup$ – user2345215 Nov 2 '14 at 2:41
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If $n=2k+1$, then $$\int_{\pi/2}^\pi\frac{\cos((2k+1)\theta)}{b^2-a^2\cos(2\theta)}d\theta=\int_0^{\pi/2}\frac{\cos((2k+1)(\pi-\theta))}{b^2-a^2\cos(2(\pi-\theta))}d\theta=\int_0^{\pi/2}\frac{-\cos((2k+1)\theta)}{b^2-a^2\cos(2\theta)}d\theta$$ and the integral is $0$ because the convergence is trivial. So assume $n=2k$, then $$\int_0^\pi\frac{\cos(2k\theta)}{b^2-a^2\cos(2\theta)}d\theta=\int_0^{2\pi}\frac{\cos(k\theta)}{2b^2-2a^2\cos\theta}d\theta=\frac1{a^2}\int_0^{2\pi}\frac{\cos(k\theta)}{2c-2\cos\theta}d\theta,$$ where $c=b^2/a^2>1$. Now \begin{align*}\int_0^{2\pi}\frac{\cos(k\theta)}{2c-2\cos\theta}d\theta&=\text{Re}\int_0^{2\pi}\frac{e^{ik\theta}}{2c-(e^{i\theta}+e^{-i\theta})}d\theta=\text{Re}\frac1i\int_0^{2\pi}\frac{-e^{ik\theta}\cdot ie^{i\theta}}{e^{2i\theta}-2ce^{i\theta}+1}d\theta\\&=\text{Re}\frac1i\int_\varphi\frac{-z^k}{z^2-2cz+1}dz,\end{align*} where $\varphi(\theta)=e^{i\theta}$ on $[0,2\pi]$. The denominator has 2 roots and only $\gamma=c-\sqrt{c^2-1}$ is inside $\text{Int}\,\varphi$, the residue is $$\frac{-\gamma^k}{2\gamma-2c}=\frac{(c-\sqrt{c^2-1})^k}{2\sqrt{c^2-1}}.$$ Putting all things together yields the answer $$\frac1{a^2}\text{Re}\frac{2\pi i}i\frac{(c-\sqrt{c^2-1})^k}{2\sqrt{c^2-1}}=\frac{\pi(c-\sqrt{c^2-1})^k}{a^2\sqrt{c^2-1}}=\frac{\pi\Big(b^2-\sqrt{b^4-a^4}\Big)^k}{a^{2k}\sqrt{b^4-a^4}}$$

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  • $\begingroup$ Yeah, right, I forgot about residue theorem used to calculate these sort of integrals. $\endgroup$ – MathematicalPhysicist Nov 2 '14 at 7:04

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