Fourier cosine series on $[0,\infty)$

Question

A function $f:[0,\pi]\to\mathbb{R}$ can be written as $$f(x)=\frac{a_0}{2}+\sum_{k=1}^\infty a_k\cos(kx),$$ where $$a_k=\frac{2}{\pi}\int_0^\pi f(x)\cos(kx)dx.$$

A substitution can shift the domain $[0,\pi]$ to any arbitrary finite interval.

Suppose we consider a function $f:[0,\infty)\to\mathbb{R}$ with unbounded domain. Can we find a cosine series for $f$ on this non-compact domain (we may assume that $f$ has exponential decay)?

Answer 1

It is not clear from your question what you are trying to do. I think you have three options:

1. Use a change of variable such as $\theta = 2 \tan^{-1} x$ which transforms $[0,\infty)$ to $[0,\pi)$. Then for a function $f:[0,\infty)$ you could obtain the cosine series for $f(\tan(\theta/2))=a_0/2 + \sum a_n \cos n \theta$. This results in a cumbersome expansion, $$f(x) = \frac{a_0}{2} + \sum_n a_n \cos (2n \tan^{-1} x),$$ where the expression for coefficients is, $$ a_n = \frac{2}{\pi}\int_0^\infty f(\tan \theta / 2) d\theta .$$ Any other one to one function that maps $[0,\infty)$ onto $[0,\pi)$ would also work.

2. Settle for a large but finite interval using the method you already described.

3. Use another form of expansion tailored to an infinite domain. One such is obtained using the Laguerre polynomials, that form a basis on $[0,\infty)$. There is plenty of information on these. They can be used to create an infinite convergent series for suitably behaved functions $f$.

Answer 2

Fourier considered the integral representation to be the limit of a discrete series as the period tended to $\infty$. The Fourier cosine version is $$ f(x) \sim \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(y)\cos(sy)dy\right) \cos(sx) ds $$ This form is correct. It can be derived from the exponential form after extending $f$ to an even function on $\mathbb{R}$. The corresponding Parseval identity is $$ \int_0^{\infty}|f(x)|^2=\frac{2}{\pi}\int_0^{\infty}\left|\int_0^{\infty}f(y)\cos(sy)dy\right|^2ds. $$