How find this $\int_{0}^{\pi}\frac{\cos{(nx)}}{\cos{x}+a}dx$

Question

Fin the integral $$I_{n}=\int_{0}^{\pi}\dfrac{\cos{(nx)}}{\cos{x}+a}dx$$ where $n\in {\mathbb N}\,,\ a>1$

My try: let $$I_{n}-I_{n-1}=\int_{0}^{\pi}\dfrac{\cos{(nx)}-\cos{(n-1)x}}{\cos{x}+a}dx$$ and note $$\cos{x}-\cos{y}=-2\sin{\dfrac{x+y}{2}}\sin{\dfrac{x-y}{2}}$$ so $$I_{n}-I_{n-1}=-2\int_{0}^{\pi}\dfrac{\sin{(nx-\dfrac{x}{2})}\sin{\dfrac{x}{2}}}{\cos{x}+a}dx$$

Then I can't ,Thank you for your help.

Answer 1

A little complex analysis helps. First, by the evenness of $\cos$, we have

$$I_n = \frac12 \int_{-\pi}^\pi \frac{\cos (nx)}{a+\cos x}\,dx,$$

and by Euler's formula, we have $\cos (nx) = \operatorname{Re} e^{inx}$, so

$$I_n = \frac12 \operatorname{Re} \int_{-\pi}^\pi \frac{e^{inx}}{a+\cos x}\,dx.$$

Since the imaginary part is odd, the integral is real, and we can omit the $\operatorname{Re}$.

Now we write $z = e^{ix}$, then we have

$$\cos x = \frac12(z+z^{-1});\quad e^{inx} = z^n;\quad dx = \frac{dz}{iz};$$

so we obtain

$$I_n = \int_{\lvert z\rvert = 1} \frac{z^n}{2a + z + z^{-1}}\, \frac{dz}{iz} = 2\pi\cdot\frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \frac{z^n}{z^2 + 2az + 1}\,dz.$$

We need the zeros of the denominator $z^2 + 2az + 1 = (z+a-\sqrt{a^2-1})(z+a+\sqrt{a^2-1})$ to apply the Cauchy integral formula. Of the zeros, $-a+\sqrt{a^2-1}$ lies inside the unit disk, and $-a-\sqrt{a^2-1}$ outside the closed unit disk. With $f(z) = \frac{z^n}{z+a+\sqrt{a^2-1}}$, the Cauchy integral formula yields

$$\begin{align} I_n &= 2\pi\cdot\frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \frac{f(z)}{z+a-\sqrt{a^2-1}}\,dz\\ &= 2\pi\cdot f(-a+\sqrt{a^2-1})\\ &= \frac{\pi}{\sqrt{a^2-1}}(\sqrt{a^2-1}-a)^n.\tag{1} \end{align}$$