# Ngô Quốc Anh

## September 8, 2007

### “Tích phân xác định” vs. “Chuỗi số”

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:20

Đề bài. $\forall\,n\in\Bbb{N}^{+}$ hãy tính

$\int_{-\pi}^{\pi}\;\;\frac{\cos\,(n\cdot x)}{5+4\,\cos\,(x)}\;\;\textbf dx$

Cách giải 1. (sử dụng giải tích phức)Instead of finding

$\int_{-\pi}^{\pi}\frac{\cos nx}{5+4\cos x}dx$

rather find

$\int_{-\pi}^{\pi}\frac{e^{inx}}{5+4\cos x}dx$

If $|z|=1$ is a unit circle from $-\pi$ to $\pi$ we have that,

$\int_{-\pi}^{\pi}\frac{e^{inx}}{5+4\cos x}dx =\oint_{|z|=1}\frac{z^{n}}{5+2(z+z^{-1})}\cdot\frac{1}{iz}dz$

Which works out to be,

$\frac{1}{i}\oint_{|z|=1}\frac{z^{n}}{(2z+1)(z+2)}dz$

This contour integral has only a single pole within $|z|=1$ which is simple $z=-1/2$.
So,

$\mbox{res}(-1/2) =\lim_{z\to-1/2}(z+1/2)\cdot\frac{z^{n}}{(2z+1)(z+2)}=\frac{\left(-1/2\right)^{n}}{3}=\frac{(-1)^{n}}{3\cdot 2^{n}}$

By residue theorem together with the $1/i$ factor gives us,

$\int_{-\pi}^{\pi}\frac{\cos nx}{5+4\cos x}dx = (-1)^{n}\frac{\pi}{3\cdot 2^{n-1}}$

Cách giải 2. (sử dụng sai phân) Let

$I_{n}=\int_{-\pi}^{\pi}\frac{\cos nx}{5+4\cos x}\, dx$.

Using some trigonometry, $\cos (n+1)x+\cos (n-1)x = 2\cos nx\cos x$ and

$I_{n+1}+I_{n-1}\; =\;\int_{-\pi}^{\pi}\frac{2\cos nx\cos x}{5+4\cos x}\, dx\; =\;-\frac{5}{2}I_{n}$

Two roots of $2x^{2}+5x+2 = 0$ is

$x =-2\text{ or }-\frac{1}{2}$.

Then we can write

$I_{n}= A (-2)^{n}+B\left(-\frac{1}{2}\right)^{n}$.

But since $I_{n}$ is bounded, $A = 0$ and

$I_{n}\; =\; I_{0}\left(-\frac{1}{2}\right)^{n}\; =\;\frac{2\pi}{3}\left(-\frac{1}{2}\right)^{n}$.

Cách giải 3. (sử dụng chuỗi số) For

$|b| < 1\quad\sum_{k = 0}^{\infty}b^{k}\cos kx =\frac{1-b\cos x}{1-2b\cos x+b^{2}}\;\Rightarrow\; 2\sum_{k = 0}^{\infty}b^{k}\cos kx-1 =\frac{1-b^{2}}{1-2b\cos x+b^{2}}$

$\int_{-\pi}^{\pi}\frac{\cos nx}{1-2b\cos x+b^{2}}=\frac{1}{1-b^{2}}\int_{-\pi}^{\pi}\cos nx\left(2\sum_{k = 0}^{\infty}b^{k}\cos kx-1\right) dx$

$=\frac{1}{1-b^{2}}\left[2\sum_{k = 0}^{\infty}b^{k}\underbrace{\int_{-\pi}^{\pi}\cos nx\,\cos kx\, dx}_{\pi\,\delta_{nk}}-\underbrace{\int_{-\pi}^{\pi}\cos nx dx}_{ = 0}\right] =\frac{2\pi\, b^{n}}{1-b^{2}}$

In particular for

$b =-\frac{1}{2}\quad\int_{-\pi}^{\pi}\frac{\cos nx}{1+\cos x+\frac{1}{4}}\, dx =\frac{2\pi}{1-\frac{1}{4}}\left(-\frac{1}{2}\right)^{n}$