Since the coefficients

$$a_k = \frac1{2\pi i}\oint_C\frac{f(z)}{(z-c)^{k+1}}\,dz$$

for the Laurent series

$$f(z)\Big|_{r\le|z|\le R} = \sum_{k=-\infty}^{\infty}a_k\cdot(z-c)^k $$

of a function $f\in\mathcal H(B(r,R))$ (i.e. a function that is holomorphic on the Annulus of radii $r\le R$), evaluated on a circle of radius $\rho\in[r,R]$,

$$\tilde a_k := \rho^k a_k = \frac1{2\pi}\int\limits_{\phi=0}^{2\pi} f(c+\rho e^{i\phi}) e^{-ik\phi}\,d\phi \\\Rightarrow f(c+\rho e^{i\phi}) = \sum_{k=-\infty}^\infty \tilde a_k\,e^{ik\phi},$$

are (up to the factor $\rho^k$) equivalent to the Fourier coefficients of a function

$$\tilde f(x):=f\Big(c+\rho e^{i\tfrac x{2\pi\rho}}\Big),\quad x\in[-\pi\rho,+\pi\rho],$$

I was wondering if there is a meaningful limit $\rho\to\infty$ for the Laurent series. For the Fourier series, the limit becomes the Fourier transform with continous "coefficients", $\tilde a_k\to \tilde a(k)\equiv \mathcal F\{\tilde f(x)\}(k)$, such that

$$\tilde f(x) = \int_{-\infty}^\infty \tilde a(k) e^{ikx}\, dk, \\\tilde a(k) = \frac1{2\pi}\int_{-\infty}^\infty \tilde f(x) e^{-ikx}\, dx.$$

So in a similar way, this limit would turn the Laurent series into a "Laurent transform",

$$f(z)\Big|_{r\le|z|} = \int_{-\infty}^\infty a(k)\cdot (z-c)^k, \\ a(k) = \lim_{R\to\infty} \frac1{2\pi}\oint\limits_{|z|=R}\frac{f(z)}{(z-c)^{k+1}}\,dz$$

(up to some factors that I probably forgot, and assuming that $f$ remains holomorphic for $R\to\infty$), describing $f(z)$ by its values at $\mathbb C$-infinity.

So, does this make sense, has it been investigated upon before and/or any applications?