The Fourier transform of a function is given by $$f(ξ) = \int_{-∞}^{∞} f(x) e^{-2πixξ}dx$$
the paper I was reading from says that the test function $e^{2πixξ}$ is a periodic with period $2\pi/ξ$ and it continues by saying ("so integrating $f$ against this test function gives information about the extent to which this frequency occurs in $f$")
My question is how integrating our original function with this "test function" gives us information about the extent to which frequency occurs in $f$ ?
On
The Fourier transform of a periodic function $f(x)$ is related to its frequency content since
$$\mathcal{F}_x\left[a_n \cos\left(2 \pi \frac{n}{P} x\right)\right](\omega )=\int_{-\infty}^{\infty} a_n \cos\left(2 \pi \frac{n}{P} x\right)\, e^{-2 i \pi \omega x} \, dx\\=\frac{a_n}{2}\, \delta\left(\omega-\frac{n}{P}\right)+\frac{a_n}{2}\, \delta\left(\frac{n}{P}+\omega\right)\tag{1}$$
and
$$\mathcal{F}_x\left[b_n \sin\left(2 \pi \frac{n}{P} x\right)\right](\omega)=\int\limits_{-\infty}^{\infty} b_n \sin\left(2 \pi \frac{n}{P} x\right)\, e^{-2 i \pi \omega x} \, dx\\=\frac{b_n}{2} i\, \delta\left(\frac{n}{P}+\omega\right)-\frac{b_n}{2} i\, \delta\left(\omega-\frac{n}{P}\right)\tag{2}$$
where $\delta(\omega)$ is the Dirac delta function, but the Fourier transform of a periodic function doesn't converge in the usual sense, rather it converges only in a distributional sense.
Its seems to me the notion that the Fourier transform of a non-periodic function $f(x)$ somehow gives the frequency content of the function $f(x)$ analogous to the coefficients of a Fourier series is perhaps a bit misguided since for a non-periodic function $f(x)$ I believe one has
$$c_n=\lim\limits_{P\to\infty}\left(\frac{1}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)\, e^{-i 2 \pi \frac{n}{P} x} \, dx\right)=0\tag{3}.$$
For example consider the function
$$f(x)=e^{-\pi^2 x^2}\tag{4}$$
with Fourier transform
$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\int\limits_{-\infty}^{\infty} e^{-\pi^2 x^2}\, e^{-i 2 \pi \omega x} \, dx=\frac{e^{-\omega^2}}{\sqrt{\pi}}\tag{5}.$$
Now since $f(x)=e^{-\pi^2 x^2}$ is an even function of $x$ consider the Fourier $\cos$ series coefficient
$$a_n=\frac{2}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x) \cos\left(2 \pi \frac{n}{P} x\right) \, dx=\frac{2}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} e^{-\pi^2 x^2} \cos\left(2 \pi \frac{n}{P} x\right) \, dx\\=\frac{i e^{-\frac{n^2}{P^2}}}{\sqrt{\pi} P} \left(\text{erfi}\left(\frac{n}{P}-\frac{i \pi P}{2}\right)-\text{erfi}\left(\frac{n}{P}+\frac{i \pi P}{2}\right)\right)\tag{6}$$
where the function $f(x)=e^{-\pi^2 x^2}$ can be approximated on the interval $-\frac{P}{2}<x<\frac{P}{2}$ by the Fourier $\cos$ series
$$f(x)\approx\frac{a_0}{2}+\lim\limits_{N->\infty}\left(\sum_{n=1}^N a_n \cos\left(2 \pi \frac{n}{P} x\right)\right)\tag{7}.$$
For the function $f(x)=e^{-\pi^2 x^2}$ note that
$$\underset{P\to\infty}{\text{lim}}a_n=\underset{P\to\infty}{\text{lim}}\left(\frac{i e^{-\frac{n^2}{P^2}}}{\sqrt{\pi} P} \left(\text{erfi}\left(\frac{n}{P}-\frac{i \pi P}{2}\right)-\text{erfi}\left(\frac{n}{P}+\frac{i \pi P}{2}\right)\right)\right)=0\tag{8}.$$
This is consistent with Parseval's theorem which can be stated as
$$\frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x)^2 \, dx=\frac{1}{2} (a_0)^2+\sum\limits_{n=1}^\infty \left(a_n^2+b_n^2\right)\tag{9}$$
for a $2 \pi$-periodic representation of a function $f(x)$ and which I believe can be generalized as
$$\frac{2}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)^2 \, dx=\frac{1}{2} (a_0)^2+\sum\limits_{n=1}^\infty \left(a_n^2+b_n^2\right)\tag{10}$$
for a $P$-periodic representation of a function $f(x)$ where $a_n$ and $b_n$ are the coefficients of the related $\cos$ and $\sin$ terms in the Fourier series representation of $f(x)$.
For the function $f(x)=e^{-\pi ^2 x^2}$ one has
$$\int_{-\infty }^{\infty } \left(e^{-\pi ^2 x^2}\right)^2 \, dx=\frac{1}{\sqrt{2 \pi }}\tag{11}$$
which implies for the function $f(x)=e^{-\pi ^2 x^2}$ one has
$$\lim_{P\to\infty}\left(\frac{2}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)^2 \, dx\right)=\frac{1}{2} (a_0)^2+\sum\limits_{n=1}^\infty \left(a_n^2+b_n^2)\right)=0\tag{12}$$
which implies $a_n=b_n=0$.
Finally, the magnitude of the frequency $\omega_n=\frac{n}{P}$ in the Fourier $cos$ series of an even function $f(x)$ related to the term in formula (1) above is not the value of $F\left(\omega_n\right)$ where $F(\omega)$ is the Fourier transform of $f(x)$, but rather the "energy" at $F\left(\omega_n\right)$ which for an even function $f(x)$ can be calculated as
$$\lim\limits_{\epsilon\to 0^+}\left(2 \int_{\omega_n-\epsilon}^{\omega_n+\epsilon} F\left(\omega-\omega_n\right) \, d\omega\right)\tag{13}$$
which for $F(\omega)=\frac{e^{-\omega^2}}{\sqrt{\pi}}$ in formula (5) above evaluates to zero which is consistent with formulas (3), (8), and (12) above.
Actually, the definition of the Fourier transform, namely $$ \hat{f}(\xi) = \int_\Bbb{R} f(x)e^{-2\pi ix\xi} \,\mathrm{d}x, $$ corresponds to a (hermitian) inner product $\langle f, e^{2\pi ix\xi} \rangle$, where $$ \langle \phi,\psi \rangle := \int_\Bbb{R} \phi(x)\overline{\psi(x)} \,\mathrm{d}x. $$ In the case of Fourier series, the trigonometric "monomials" $e^{2\pi ix\xi_n}$, with $\xi_n = \frac{n}{T}$, form an orthonormal basis of the Lebesgue space $L^2([-\frac{T}{2},\frac{T}{2}])$, which is a functional vector space for recall, in such a way that the expression $$ f(x) = \sum_n c_n e^{2\pi ix\xi_n} $$ is nothing else than the projection of $f$ onto the Fourier basis. In consequence, the coefficient $c_n = \langle f, e^{2\pi ix\xi_n} \rangle$ "measures" how much $f$ "contains" the sinusoid $e^{2\pi ix\xi_n}$ as its coordinate in that particular basis. It is to be noted that the Taylor series are actually constructed in the same manner with respect to a polynomial basis.
Also, note that this formalism was developped to study waves historically, that is why one speaks of the amplitude associated to the frequency $\xi_n$ when referring to $c_n$, even if those quantities have to be understood in an abstract sense and might carry other physical units.
Finally, mutatis mutandis, the Fourier transform can be interpreted in the same way, except for the fact that the trigonometric functions $e^{2\pi ix\xi}$ are meant to form the continuous basis of a larger space.