Prove $\cos(ax)=\frac{\sin(\pi a)}{\pi a}+\sum_{n=1}^{\infty }(-1)^{n}\frac{2a\sin(\pi a)}{\pi (a^{2}-n^{2})}\cos(nx)$

Question

Prove that for all $-\pi\le x\le \pi$ and all $a \notin \mathbb Z$, $$\cos(ax)=\frac{\sin(\pi a)}{\pi a}+\sum_{n=1}^{\infty }(-1)^{n}\frac{2a\sin(\pi a)}{\pi (a^{2}-n^{2})}\cos(nx).$$

I understand general Fourier analysis; however, I am unsure of any steps I should follow to be able to solve this sort of question. Any help or links to material that would help me to understand the process of how to solve a question like this would be much appreciated.

Answer 1

Hint:

I leave the details to the OP (integrability issues and the like)

Consider the function $f_a(x)=e^{a ix}$ over $(-\pi,\pi)$ for $a\in(0,1)$. Its Fourier coefficients are given by $$c_n(f_a)=\frac1{2\pi}\int^{\pi}_{-\pi}e^{i(a-n)x}\,dx=\sin(a\pi)\frac{(-1)^n}{\pi(a-n)}$$

From this, it follows (why?) that

$$f_a(x)=e^{a ix} =\sum_{n\in\mathbb{Z}}c_n(f_a) e^{in x}= \sin(a\pi)\sum_{n\in\mathbb{Z}} \frac{(-1)^n}{\pi(a-n)}e^{inx}$$ almost surely (in fact for all points in $(-\pi,\pi)$ by continuity and differentiability of $f_a$). Matching real and imaginary parts gives

$$\cos(a x)=\sin(a\pi)\sum_{n\in\mathbb{Z}} \frac{(-1)^n}{\pi(a-n)}\cos(nx)$$

Now split the sum $\sum_{n\in\mathbb{Z}}T_n(x)=T_0(x)+\sum^\infty_{n=1}(T_n(x)+T_{-n}(x))$ where $T_m(x)=\frac{(-1)^m}{\pi(a-m)}\cos(mx)$ for all $m\in\mathbb{Z}$.

Comment: A similar expression can be obtained for $\sin(a x)$.

Answer 2

For function $f$, continuous on $[-\pi,\pi],$ the Fourier series takes the form $$\hspace{4cm}f(x)\sim {a_0\over 2}+\sum_{n=1}^\infty (a_n\cos nx +b_n\sin nx) \qquad \qquad \qquad\qquad (*)$$ where $\hspace{2cm}$ $\displaystyle a_n={1\over \pi}\int\limits_{-\pi}^\pi f(x) \cos nx\,dx,\qquad b_n={1\over \pi}\int\limits_{-\pi}^\pi f(x) \sin nx\,dx$

At any point $-\pi<x<\pi,$ where for example $f(x)$ is differentiable, or satisfies the Lipischitz condition, the series is convergent and we have equality in $(*)$ . The continuity at $x$ is not sufficient, i.e. the series might not converge. At the endpoints $\pm\pi$ the series is convergent and equal to ${1\over 2}[f(-\pi)+f(\pi)] ,$ when the function $f$ has one sided derivatives at these points, or satisfies one sided Lipschitz conditions there.

The function $f(x)=\cos ax$ satisfies $f(-x)=f(x),$ hence $b_n=0$ for any $n.$ Next $$\pi a_n=\int\limits_{-\pi}^\pi \cos ax \cos nx\,dx=2\int\limits_{0}^\pi \cos ax \cos nx\,dx \\ = \int\limits_0^\pi [\cos(n+a)x+\cos (n-a)x]\,dx ={\sin(n+a)x\over n+a}+{\sin (n-a)x\over n-a}\Big |_0^\pi \\ ={\sin (n\pi+\pi a)\over n+a }+{\sin (n\pi-\pi a)\over n-a }=(-1)^n{2a\sin \pi a\over a^2-n^2}$$ Summarizing $${a_0\over 2}={\sin \pi a\over \pi a},\qquad a_n=(-1)^n{2a\sin \pi a\over \pi(a^2-n^2)}$$ and the equality in $(*)$ holds for $-\pi<x<\pi.$ The equality $(*)$ at the endpoints does not hold, as $e^{-\pi a}\neq e^{\pi a}.$ Instead we have $${1\over 2}[e^{-\pi a}+ e^{\pi a}]={\sin \pi a\over \pi a}+\sum_{n=1}^\infty (-1)^n{2a\sin \pi a\over \pi(a^2-n^2)}\cos n\pi={\sin \pi a\over \pi a}+\sum_{n=1}^\infty {2a\sin \pi a\over \pi(a^2-n^2)}$$