Fourier transform of a normalized function

Question

If I have a normalized function (ie $\int_\infty^\infty f(x)dx=1$), and then apply a fourier transform to the function, it does not appear to generally be true that the fourier transform is normalized. I am aware of Parseval's theorem/Plancherel's theorem that the norm squared of the function and the fourier transform have the same value when integrated over the real line, but am I correct there is no such theorem for simply the integral of the whole function over the reals? I have some vague memory that I could arrive at a normalization by multiplying/diving by a factor of $2\pi$ but at the moment I think this is just me mixing up the fourier transform normalization constant.

Answer 1

Note that

$$\begin{align*}\int^\infty_{-\infty} f(x) dx &= \int^\infty_{-\infty} f(x) e^{-2\pi i sx} dx \biggr\rvert_{s=0} \\ \\ &= \mathscr{F}\left\{f(x)\right\}(0) \\ \end{align*}$$

By enforcing

$$\begin{align*}\int^\infty_{-\infty} f(x) dx &=1 \\ \end{align*}$$

all that you have really enforced is that the DC component, or mean, of the function is $1$.

So the following is not really a theorem, but more just a property of the Fourier Transform:

The Fourier Transform evaluated at frequency $0$ is the integral of the whole function over the reals.