My consideration might be total nonsense (as a high school student, I lack the mathematical knowledge to really check my idea), but I was just wondering whether one could find a continuous generalization of the Taylor series comparable to the continuous Fourier transform.
Just like the Fourier transform decomposes a function $f(t)$ into a sum of trigonometric functions with a result $\hat{f}(\omega)=\mathcal{F}(f)(\omega)$ with the domain of frequencies, could one define a Taylor transform $\mathcal{T}(f)$ that would operate on $n$th derivatives?
Instead of the trigonometric functions in Fourier transform, $f$ would be decomposed into a sum of polynomials of form $\frac{1}{n!}x^n$.
So is it a valid operation to generalize the Taylor series for continuous values of $n$ (i.e. using integrals instead of sums)? My naive approach would be
$$f(t) = \intop_0^{\infty} \mathcal{T}(f)(n)\cdot \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$
So in fact, the transform would define (or at least require) non-integer derivatives of a function. Do these exist (like a non-integer iterates of functions can be defined too)?
Example: Assuming that $\frac{\mathrm{d}^ne^x}{\mathrm{d}x^n}(0) = 1$ for any $x\in \mathbb{R}$, I'd come up with
$$e^t = \intop_0^{\infty} \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$ as a continuous version of
$$e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!}$$
Thus $$\mathcal{T}(e^t)(n) = 1,\,n\in\mathbb{R}$$
Is a transform like I explained possible or - in case it is - even somehow useful?