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

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.

• Whether the formal power series yields a well-defined operator will depend on the function space you're working with; a sufficient pair of conditions for the series to be well-defined is for your space to be a Banach space on which $I$ is bounded. These properties guarantee that the partial sums of the series are a Cauchy sequence in a complete space. One concrete example of such a space are continuous functions on a compact interval under the supremum norm. I don't know whether $e^I$ is interesting so I'll leave that for someone else. Jun 29 at 6:11

$$e^I$$ 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 $$\begin{gather*} x^2I_1''(x)+xI_1'(x)-(1+x^2)I_1(x)=0 \\ I_1(0)=0 \end{gather*}$$