The Fourier series coefficients are often derived by assuming a function can be represented as a series
$$f(x) = \sum_{n=0}^\infty A_n \cos\left(\frac{2\pi n x}{L}\right) + \sum_{n=0}^\infty B_n \sin\left(\frac{2\pi n x}{L}\right)$$
then multiplying by $\sin(2\pi m x / L)$ or $\cos(2\pi m x / L)$ and integrating over $[-L, L]$, exploiting the orthogonality of of these trig functions over $[-L,L]$ to show that
$$A_n = \frac{1}{L}\int_{-L}^L f(x) \cos\left(\frac{2\pi n x}{L}\right)dx$$
For the Fourier transform, we can try to do the same thing, assuming we can represent our function as an integral over trig functions of continuous frequencies, i.e.
$$f(x) = \int_{-\infty}^\infty A_n(\xi) e^{2\pi i x \xi} d\xi$$
And then multiplying by $e^{-2\pi i x k}$ and integrating
$$\int_{-\infty}^{\infty} f(x)e^{-2\pi i x k} dx = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} A(\xi) e ^{2\pi i x(\xi – k)} d\xi dx.$$
The left-hand side is clearly the Fourier transform formula, so I'm assuming I can reduce the right-hand side to (some scalar multiple of) $A(k)$, but haven't been able to see a way forward. The Riemann-Lebesgue lemma gives us something, but I'm wondering if there is any orthogonality trick over the whole real line $\mathbb{R}$ which lets simplify the right-hand side and get an explicit formula for $A(k)$, i.e. $\hat{f}(\xi)$.
Best Answer
Final Edit: Of course the result you're looking for is just the $L^1$ Inversion Theorem. Below there are some fuzzy thoughts about how IT might be derived from Fourier series - see here for a version of what's below that's an actual proof!
Edit: Thinking about what I said the other day I realized it explains something about the Fourier transform that's always been somewhat mysterious to me. So I'm suddenly enthusiastic about all this. Happens all the time that I learn things by answering MSE questions, but usually regarding things like algebra that I know nothing about, not things I sort of understand, like Fourier analysis. See below...
Original Answer:
Not quite exactly what you're asking for, but it seems to me you should be able to derive Fourier inversion from Fourier series via Poisson summmation. For sufficiently well-behaved functions; this is doubtless going to give a much weaker result than the standard inversion theorem.
Assuming you'd like to work out the details for yourself: If $f\in L^1(\Bbb R)$ and $L>0$ define $$f_L(t)=\sum_{k\in\Bbb Z}f(t+kL).$$Then $f_L$ has period $L$, so it has a Fourier series. Haven't worked it out in detail but it seems clear that (under suitable hypotheses) if you say $f_L$ equals its Fourier series and then let $L\to\infty$ it should follow that $f$ is the inverse tranform of $\hat f$.
Edit, explaining the above a little more explicitly:
Note first that nothing below this line is actual math, quite. The hypotheses are missing - we assume that everything always converges to what it "should" converge to...
For $f\in L^1(\Bbb R)$ define the Fourier transform $\hat f$ by $$\hat f(\xi)=\int f(t)e^{-it\xi}\,dt.$$(Any time you're talking about the Fourier transform you should really include the definition, even in a context where the reader certainly knows the definition, because everyone puts the $\pi$'s in different places; if the reader's definition is a little different things won't look right. This is one reason for the Littlewood Convention, to the effect that $2\pi=1$.)
We're after
Something that Euler or Fourier might have regarded as a proof:
Define $f_L$ as above. Then $f_L$ has period $L$. In our current fantasy periodic functions are always equal to the sum of their Fourier series, so $$f_L(t)=\sum_nc_{L,n}e^{2\pi i nt/L},$$where $$c_{L,n}=\frac1L\int_0^Lf_L(t)e^{-2\pi int/L}.$$Now if you insert the definition of $f_L$ and note that that exponential has period $L$ you see that $$c_{L,n}=\frac1L\hat f\left(\frac {2\pi n}L\right),$$so we have $$f_L(t)=\frac 1L\sum_n\hat f\left(\frac {2\pi n}L\right)e^{2\pi i nt/L}.$$
But $\frac {2\pi}L\sum_n\hat f\left(\frac {2\pi n}L\right)e^{2\pi i nt/L}$ is precisely a Riemann sum for $\int\hat f(\xi)e^{i\xi t}\,d\xi$; since $f_L(t)\to f(t)$ as $L\to\infty$ the theorem follows.
Here's why this seems so cool to me, even though it's really not quite an actual proof: I know the standard proof, or a standard proof, of the Inversion Theorem very well. It's never been intuitively clear to me where the $2\pi$ comes from - some integral has some value, if that integral were different it would be a different constant. But here it's really obvious why the $1/2\pi$ is there: The constants for Fourier series are clear, just because of orthogonality, and the Fourier transform simply inherits the $1/2\pi$ from Fourier series. Ahh, that's better.
So it seems like a worthwhile project to try to concoct an actual proof of IT more or less as above. I can imagine at least two approaches: (i) Show that the argument works for $f_n$, where $f_n\to f$ almost everywhere and $||\hat f_n-\hat f||_1\to0$, (ii) show that the argument actually works assuming just $f,\hat f\in L^1$.
(Of course if we're attempting (ii) we can't show that the Fourier series for $f_L$ converges to $f_L$, since that's simply false in general. But the Fourier series is "summable" to $f_L$...)
Edit: In fact it turns out (i) is not hard - see here. Briefly, assuming $f,f',f''\in L^1$ is enough to make the argument above work, and deriving the full Inversion Theorem from this special case is easy.