Fourier cosine series on $[0,\infty)$ 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)?
 A: It is not clear from your question what you are trying to do.  I think you have three options:

*

*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.


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


*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 $f$.
A: 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
$$
          \int_0^{\infty}|f(x)|^2=\frac{2}{\pi}\int_0^{\infty}\left|\int_0^{\infty}f(y)\cos(sy)dy\right|^2ds.
$$
