Is there a well-defined exponential of an integral operator?

Question

I’ve seen $e^{\frac{d}{dx}}$ yield the shift operator, but would $e^{I}$ where $If(x)\equiv \int f(x)dx$ yield something interesting as well (or even something at all)? I’d imagine that it’s defined as the integral operator is better behaved than the differentiation operator, but I it interesting and does this operator have a name?

$e^I$ is just shorthand for $\sum_{n=0}^{\infty} \frac{I^n}{n!}$ as is oftentimes done in operator theory and related fields.

Answer 1

$e^I$ has a nice expression in terms of an integral kernel, but I don't know of any uses for it.

Rather than work with quotient vector spaces, let $$(If)(x)=\int_0^x{f(t)\,dt}$$ Then, for all $n\in\mathbb{Z}^+$, $$(I^nf)(x)=\int_0^x{\frac{(x-t)^{n-1}}{(n-1)!}f(t)\,dt}$$ a result due to Cauchy and easy to prove by induction.

Applying Cauchy's formula and noting that power series converge uniformly, \begin{align*} (e^If)(x)&=f(x)+\sum_{n=1}^{\infty}{\frac{1}{n!}\int_0^x{\frac{(x-t)^{n-1}}{(n-1)!}f(t)\,dt}} \\ &=f(x)+\int_0^x{f(t)\sum_{n=1}^{\infty}{\frac{(x-t)^{n-1}}{(n-1)!n!}}\,dt} \\ &=f(x)+\int_0^x{\frac{I_1(2\sqrt{x-t})}{\sqrt{x-t}}f(t)\,dt} \end{align*} where $I_1$ solves the ODE \begin{gather*} x^2I_1''(x)+xI_1'(x)-(1+x^2)I_1(x)=0 \\ I_1(0)=0 \end{gather*}