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}

 

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