Show $\int_{-\pi}^{\pi} \frac{1 - \cos (n+1) x}{1- \cos x} dx = (n+1) 2 \pi$ for $n \in \mathbb N$ [duplicate]

Possible Duplicate:
Compute the trigonometric integrals

For $n \in \mathbb N$, $$\int_{-\pi}^{\pi} \frac{1 - \cos (n+1) x}{1- \cos x} dx = (n+1) 2 \pi$$

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marked as duplicate by J. M., Gerry Myerson, Did, t.b., NorbertJul 27 '12 at 16:24

What did you try? (Unrelated: do you plan to leave the mess there as is?) – Did Jul 27 '12 at 9:30
@did I have tried $\sin^2 A = (1 - \cos 2A )/2$ but it doesn't work. – Bamily Jul 27 '12 at 9:39
OK. What else did you try? The answer to at least one of your other questions involves some manipulations of trigonometric lines. Did you try to apply these here? – Did Jul 27 '12 at 9:48
@did Oh Thank you did. Using your hint, I've shown this easily. – Bamily Jul 27 '12 at 9:55
Am I going blind, or do I not see a question mark anywhere? – J. M. Jul 27 '12 at 11:27

Maybe you wanna use this fact based upon the arithmetic progression rule: $$I_{n} = \frac{I_{n-1}+I_{n+1}}{2}$$ Let's consider the simpler case $$I_{n}=\int_{0}^{\pi}\frac{1-\cos nx}{1-\cos x} dx$$ Then

$$\frac{I_{n-1}+I_{n+1}}{2} =\int_{0}^{\pi} \frac{2-\cos(n-1)x-\cos (n+1)x}{2(1-\cos x)} dx= \int_{0}^{\pi} \frac{1- \cos nx \cos x}{1- \cos x}dx=$$ $$\int_{0}^{\pi} \frac{(1-\cos nx) + \cos nx(1-\cos x)}{1-\cos x} dx = I_{n} +\int_{0}^{\pi} \cos nx \space dx=I_{n}$$

Since the integrand is an even function we get $$2I_{n+1}=\int_{-\pi}^{\pi}\frac{1-\cos (n+1)x}{1-\cos x} dx$$ But $$I_{1}=\int_{0}^{\pi} dx=\pi$$ $$I_{2}-I_{1}=\int_{0}^{\pi} \frac{\cos x -\cos 2x}{1-\cos x} dx = \int_{0}^{\pi} 2 \cos x +1 \space dx = \pi$$ $$I_{n}=n \pi$$ Finally, we get that $$\int_{-\pi}^{\pi}\frac{1-\cos (n+1)x}{1-\cos x}=2I_{n+1}=(n+1) 2 \pi$$

Q.E.D.

I also think you wanna see this.

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Setting $z=e^{ix}$ and denoting by $C$ the unit circle, one has $$I_n:=\int_{-\pi}^\pi\frac{1-\cos(n+1)x}{1-\cos x}dx=-i\oint_Cf(z)dz,$$ where $$f(z)=\frac{z^{n+1}+z^{-n-1}-2}{z(z+z^{-1}-2)}=\frac{(1+z+\ldots+z^n)^2}{z^{n+1}}=\frac{(P_n(z))^2}{z^{n+1}}.$$ Thanks to the Residue Theorem, one gets \begin{eqnarray} I_n&=&2\pi\text{Res}(f,0)=\frac{2\pi}{n!}\lim_{z \to 0}\frac{d^n}{dz^n}\left[z^{n+1}f(z)\right]=\frac{2\pi}{n!}\lim_{z \to 0}\frac{d^n}{dz^n}(P_n(z))^2\cr &=&\frac{2\pi}{n!}\lim_{z \to 0}\sum_{k=0}^n{n\choose k}P_n^{(k)}(z)P_n^{(n-k)}(z) =\frac{2\pi}{n!}\sum_{k=0}^n{n\choose k}P_n^{(k)}(0)P_n^{(n-k)}(0). \end{eqnarray} Since $P_n(z)=1+z+\ldots+z^n$, one has $P_n^{(k)}(0)=k!$ for every $k \in \{0,\ldots, n\}$. Hence $$I_n=\frac{2\pi}{n!}\sum_{k=0}^n{n\choose k}k!(n-k)!=2\pi\sum_{k=0}^n1=2(n+1)\pi.$$

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