# Fourier Cosine Transform and Dirac Delta Function

Given an even $f \in L^2([-\pi,\pi]),$ we can write $f$ as a Fourier Cosine Series as such

$$f(x)= \sum_{n=0}^{\infty} \alpha_n \cos(nx),$$ where $$\alpha_0= \int_{- \pi}^{\pi} \frac{f(x)}{ 2 \pi} \ dx, \\ \alpha_n=\int_{- \pi}^{\pi} \frac{f(x) \cos(nx)}{ \pi} \ dx, n \in \mathbb{N}.$$ The reason why the normalization constants $\frac{1}{2 \pi}, \frac{1}{\pi}$ are in $\alpha_0, \alpha_n,$ respectively is that

$$\int_{-\pi}^{\pi} \frac{\cos(mx) \cos(nx)}{\pi} \ dx= \delta_{m,n}, m,n \in \mathbb{N}$$ and $$\int_{-\pi}^{\pi} \frac{\cos(nx)}{2\pi} \ dx= \delta_{0,n}, n \in \mathbb{N}.$$

Now suppose $f \in L^2( \mathbb{R})$ is even. We define the Cosine Transform as

$$\mathcal{F}_c(f(x))= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) \cos(xy) \ dx.$$ How can we obtain the $\frac{1}{\sqrt{2 \pi}}$ in a manner similar to that of the series ? I know it turns out to make the Fourier Inversion Formula nice, but assuming I hadn't known about the Fourier Transform Properties, how can we obtain this normalization constant ? How do we know it could not be any other value like $\frac{1}{\sqrt{4 \pi}},\frac{1}{\sqrt{6 \pi}},\frac{1}{\sqrt{8 \pi}},$ .. etc ?

I also informally made some other observations.

The structure of the transform seems to indicate it is the continuous analogue of the $\alpha_n$ used in the Cosine Series, and by that logic, it looks like the double integral $$f(x)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(z) \cos(zy) \cos(xy) \ dz \ dy,$$ is the continuous analogue of the entire series. Are these observations "correct" ?

I noticed while playing around with the cosine integral that $$\delta(m-n)= \frac{1}{ 2 \pi} \int_{-\infty}^{\infty} \cos(mx)\cos(nx) \ dx$$ if we now let $m,n \in \mathbb{R}.$ So in a sense, can the Dirac Delta Distribution can be viewed as a continuous analogue of the Kronecker Delta Function, and as before, why does the $\frac{1}{ 2 \pi}$ make things work ?

• It seems to me that there are some set-up issues: for one, the function $f$ is not equal in an $L^2$ sense to "its cosine expansion" unless it is not only $2\pi$ periodic but also even. Yes, the usual "Fourier series" can be expressed either in terms of exponentials or in terms of cosines and_sines... If there are further tacit assumptions on $f$, etc., could you make them explicit, please? This (possibly) subordinate issue pervades the rest. – paul garrett Mar 29 '17 at 23:59
• I think I have made everything explicit, as I have edited it. – Vivek Kaushik Mar 30 '17 at 1:54

Indeed, that normalizing constant is often presented as though it were a magical insight. In fact, the $2\pi$ can be inserted in various fashions into "Fourier transform" or "cosine transform" on even functions. Once we discover that a normalizing constant must occur, and what it is, the only question is where to put it.
To discover the necessary constant when we have fixed other choices (for example, $\cos(xy)$ rather than $\cos(2\pi xy)$), apply a non-normalized transform and inverse transform to something whose transform is easy to understand, such as a Gaussian $f(x)=e^{-x^2}$, and discover the constant needed to make things match up.
A slightly new issue arises, namely, that if we (reasonably-enough, as a heuristic) interchange the order of integration, we seem to find that $\int_{-\infty}^\infty \cos(mx)\,\cos(nx)\;dx$ is some multiple of Dirac's $\delta$. This is entirely correct, if interpreted properly, e.g., not as a numerical integral, but "integral/cosine transform" in an extended sense (e.g., on tempered distributions, extended by continuity).
Again, the $2\pi$ is an artifact that must appear somewhere in these formulas, as we discover. No, it is completely not obvious that your (extended-sense) integral for $\delta$ needs the $2\pi$ to be correct. That is really only discovered by looking at Fourier inversion for nicer functions (e.g., Schwartz functions).