I'm studying Fourier transforms (as opposed to series) in the context of mathematics (and not quantum mechanics) for the first time, and I'm trying to be commensurately somewhat more rigorous.
A simple example which is tripping me up is the $x(t) = \cos(\omega t)$.
I know qualitatively that because the function is periodic, it's transform should be a delta function of some sort.
So I can set up the following:
$$X(f) = \int_{-\infty}^{\infty}e ^ {-i2\pi f t}cos(\omega t)dt$$
which I can change into:
$$X(f) = \int_{-\infty}^{\infty}\frac{e ^{it(\omega - 2\pi f)} + e^{-it(\omega+2\pi f)}}{2}$$
Now, I don't need to look too hard to see that this will not converge using a typical Calc II improper variable set up (unless it will; I took Calc II ten years ago with a long gap in the middle and sometimes end up flying by the seat of my pants).
Wolfram Alpha leapfrogs the problem and turns it into this:
$$X(f) = \frac{\delta(\omega - 2\pi f) + \delta(\omega + 2\pi f)}{2}$$
But there's no explanation of how we get here.
So I would appreciate the following:
1) A (softly) rigorous explanation of that jump. I have a semester of Real Analysis under my belt, so snowballing me with Lebesgue integrals isn't going to help me. I'm familiar with the idea but the technical details are beyond me (at the moment).
2) Why do we define the transform this way if a lot (anecdotally) of these integrals fail to converge (at least in the Calc II (Reimann?) sense).
Thanks
The Fourier inversion theorem says that $X(t) = \hat{\hat X}(-t),$ where $\hat{\phantom{X}}$ stands for Fourier transform: $$\hat X(f) = \int_{-\infty}^{\infty} X(t) e^{-i 2\pi ft} dt$$
If you compute the Fourier transform of $X(t) = \frac12 ( \delta(\omega-2\pi f) + \delta(\omega+2\pi f) )$ then you can thereafter apply the inversion theorem to justify $$\mathcal F\{\frac12 (e^{it(\omega-2\pi f)} + e^{-it(\omega-2\pi f)})\} = \frac12 ( \delta(\omega-2\pi f) + \delta(\omega+2\pi f) ).$$
Fourier transforms like these are justified in the theory of distributions ($\delta$ is a distribution which is not also a function). If $u$ is a distribution then the Fourier transform of it is defined in a weak sense: $$\int \hat u(x) \ \phi(x) \ dx = \int u(x) \ \hat\phi(x) \ dx$$ Here $\phi$ is a "nice" function on which there is no problem to define the Fourier transform.