Is there a continuous Taylor/MacLaurin transform (like the Fourier transform)? 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?
 A: The Laplace transform does exactly what you're talking about. The polynomials don't necessarily come in the form ${1 \over n!} x^n$ but can be substituted into that form.
Check out this lecture by Arthur Mattuck from MIT for the details.
A: I've never thought about it as a generalization of the Taylor series, but what you're describing sounds like the Mellin transform, which is defined as
$$\hat{f}(s) = \int_0^{\infty}x^{s-1} f(x)\mathrm{d}x$$
It's used a lot in certain branches of high-energy physics theory (which is how I happen to know about it). The Wikipedia article linked above describes some other uses.
A: This is a very interesting question. I am acquainted only very little with these matters. Hopefully other people can fill you in better. There is indeed some analogy of the type you are seeking. 
The Fourier series of a periodic function captures all the information for a function(signal) over a fixed period. When you determine the Fourier coefficients, you compute suitable integrals over a fixed period and you capture all the frequency information for a certain period. Fourier transforms capture frequency domain information for non-periodic signals as well. If you look at the Fourier transform of a function, you would in general be at a loss to guess the function behavior at a particular point.
On the other hand, note that Taylor Series is giving information about a function specifically at a point. Taylor series captures very very local information.
It seems there is such a tradeoff always happening. If you want more information on the frequency domain, you have to sacrifice information on the time domain, and vice versa. On the one extreme we have the Taylor series and the other the Fourier theory. It seems there are a range of transforms called the wavelet transforms which are in between these two extremes and captures a bit of both frequency domain and time domain information without completely sacrificing one of these. It seems for such transforms for precisely formulating such a tradeoff, there is a certain uncertainty principle which is a precise theorem in Harmonic analysis.
A: as David Z says, the Mellin transform is basically what you found.
if you look at Ramanujans's master theorem on wikipedia and change the notation a bit you will find the desired relation that for an analytic function $f$ which has a (almost Taylor) expansion
$$
  f(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!} \mathcal{T}[f](n)
$$
with
$$
  \mathcal{T}[f](-s) = \int_{0}^\infty \frac{x^{s-1}}{\Gamma(s)} f(x) dx  
$$
(I am not sure, however, why they use this minus-sign convention)
