# Defining $e^{x}$ as an infinite number of integrals

I was thinking that using the Taylor expansion of $e^{x}$ we use a summation of derivatives is it possible to write it using integrals? Something along the lines of

$\int\int\int...\int dxdxdx...dx$

I think this would be off by a factor of some constant. If it is possible an explanation as to what is happening would be nice. Any guidance is appreciated.

Edit:
What I mean by infinite integrals is if I was to integrate dx over and over the result begins to look like the Taylor series for $e^{x}$ for example for three integrals we have:
$\int\int\int dxdxdx = \int\int (x + c)dxdx = \int(\frac{x^{2}}{1} + cx)dx = \frac{x^{3}}{3}+\frac{cx^{2}}{2} + d$

If this process was continued would it be the Taylor Series of $e^{x}$? Is it even correct to integrate like this?

Let $T:C(\mathbb{R})\rightarrow C(\mathbb{R})$ be an operator on the space of continuous functions on $\mathbb{R}$ such that for $f\in C(\mathbb{R})$, the function $Tf\in C(\mathbb{R})$ is defined by $$(Tf)(x) = 1+\int\limits_{0}^{x}{f(t)\text{ d}t}.$$ Then, for all $x\in\mathbb{R}$, we have $$e^x = \lim\limits_{n\rightarrow\infty}{(T^n(1))(x)}$$ where "$1$" on the RHS is the function identically equal to 1.
• @ThisPlayName What's here expands out to, more or less, saying $$e^x=1+\int_{0}^{x_0}1 + \int_{0}^{x_1} 1 + \int_{0}^{x_2} 1+\int_{0}^{x_3}1 + \ldots dx_3\,dx_2\,dx_1\,dx_0$$ where we nest infinitely many integrals this way (in a sense). – Milo Brandt Nov 23 '15 at 18:00