I am reading Rudin's Principle of Mathematical Analysis chapter 8, and I am trying to evaluate the Fourier coefficients of $\log \left(2\cos \left(\frac{x}{2}\right)\right)$, with respect to $e^{inx}$ basis, and conclude that the convergence is pointwise on the interval $(-\pi, \pi )$. We can do this in various ways, but I am trying to explicitly evaluate the Fourier coefficients by integration by parts. $$2\pi c_n=\int_{-\pi }^{\pi } \log \left(2\cos \left(\frac{x}{2}\right)\right) \cdot (\cos (nx))\,dx$$ I omitted the $\sin (nx)$ part since it vanishes.
One thing that bothers me is that $\log (2\cos (\frac{x}{2}))$ does not have a period of $2\pi $. Therefore, I cannot apply the theorem I learned about pointwise convergence since the theorem assume that the function has period of $2\pi$. Could you tell me how to evaluate the above integral using integration by parts and trigonometry identities, then how to conclude pointwise convergence even though the function does not have a period of $2\pi$?
Below is the assumption of the theorem I learned about pointwise convergence.
Assumption: f(x) has period $2\pi$, and for some x, there are constants $\delta >0$ and finite $M \in \mathbb{R}_{>0}$, such that $|f(x+t)-f(x)|\leq M|t|$ for all $t\in (-\delta, \delta )$
Thanks.
You want $$\ln\left(2\cos\left(\frac x2\right)\right)=\frac{a_0}2+\sum_{n=1}^{\infty}a_n\cos(nx)$$ I can't help you much with $$a_0=\frac1{\pi}\int_{-\pi}^{\pi}\left(\ln2+\ln\cos\left(\frac x2\right)\right)dx=2\ln2+\frac1{\pi}\int_0^{\pi}\frac{(x-\pi)\sin x}{1+\cos x}dx$$ But since $\ln\cos(x/2)=\ln\sqrt{(1+\cos x)/2}=\frac12\ln(1+\cos x)-\frac12\ln2$ we have for $n\ge1$, $$\begin{align}a_n&=\frac1{2\pi}\int_{-\pi}^{\pi}\cos nx\ln(1+\cos x)dx\\ &=\left.\frac1{2\pi n}\sin nx\ln(1+\cos x)\right|_{-\pi}^{\pi}+\frac1{2\pi n}\int_{-\pi}^{\pi}\frac{\sin nx\sin x}{1+\cos x}dx\\ &=\frac1{2\pi in}\oint_{|z|=1}\frac{\frac{z^n-z^{-n}}{2i}\frac{z-z^{-1}}{2i}}{1+\frac{z+z^{-1}}2}\frac{dz}z=\frac i{4\pi n}\oint_{|z|=1}\frac{z-1}{z+1}(z^{n-1}-z^{-n-1})dz\\ &=\frac i{4\pi n}\cdot2\pi i\left.\frac1{n!}\frac{d^n}{dz^n}\left(z^{n+1}\frac{z-1}{z+1}(z^{n-1}-z^{-n-1}\right)\right|_{z=0}\\ &=\left.-\frac1{2n}\frac1{n!}\left(\frac{2(-1)^nn!}{(z+1)^{n+1}}\right)\right|_{z=0}=\frac{(-1)^{n+1}}n\end{align}$$ Where we have let $z=e^{ix}$ and integrated counterclockwise around the unit circle in the complex $z$-plane.
EDIT: OK, so you don't have to do any integrals. $$\begin{align}\ln\left(2\cos\left(\frac x2\right)\right)&=\ln\left(2\sqrt{\frac{1+\cos x}2}\right)=\ln\sqrt{2(1+\cos x)}\\ &=\frac12\ln\left(2(1+\cos x)\right)=\frac12\ln(1+2\cos x+\cos^2x+\sin^2x)\\ &=\frac12\ln\left((1+\cos x)^2+\sin^2x\right)\\ &=\frac12\left((1+\cos x+i\sin x)(1+\cos x-i\sin x)\right)\\ &=\frac12\ln\left((1+e^{ix})(1+e^{-ix})\right)\\ &=\frac12\left(\ln(1+e^{ix})+\ln(1+e^{-ix})\right)\\ &=\frac12\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n\left(e^{inx}+e^{-inx}\right)=\sum_{k=1}^{\infty}\frac{(-1)^{n+1}}n\cos nx\end{align}$$ In particular we didn't have do do the hard integral for $n=0$ because we see directly that $a_0=0$.
EDIT: Oh, the integrals are much easier than that. If $n\ge2$ then $$\begin{align}I_n&=\int_{-\pi}^{\pi}\frac{\sin nx\sin x}{1+\cos x}dx=\int_{-\pi}^{\pi}\frac{\sin\left((n-1)x\right)\cos x\sin x+\cos\left((n-1)x\right)\sin^2x}{1+\cos x}dx\\ &=\int_{-\pi}^{\pi}\left(\sin\left((n-1)x\right)\sin x-\frac{\sin\left((n-1)x\right)\sin x}{1+\cos x}+\cos\left((n-1)x\right)(1-\cos x)\right)dx\\ &=\int_{-\pi}^{\pi}\left(-\frac{\sin\left((n-1)x\right)\sin x}{1+\cos x}+\cos\left((n-1)x\right)-\cos nx\right)dx\\ &=-I_{n-1}=\frac{(-1)^{n-1}}{(-1)^n}I_{n-1}\end{align}$$ So $$(-1)^nI_n=(-1)^{n-1}I_{n-1}=(-1)^1I_1=-\int_{-\pi}^{\pi}\frac{\sin^2x}{1+\cos x}dx=-\int_{-\pi}^{\pi}(1-\cos x)dx=-2\pi$$ And for $n\ge1$, $$a_n=\frac1{2\pi n}I_n=\frac{-(-1)^n\cdot2\pi}{2\pi n}=\frac{(-1)^{n+1}}n$$ Then we have for $x=0$, $$\begin{align}\ln\left(2\cos\left(\frac x2\right)\right)&=\ln2=\frac{a_0}2+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n\cos nx\\ &=\frac{a_0}2+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n=\frac{a_0}2+\ln2\end{align}$$ Showing that $a_0=0$.