Evaluation of $\exp\left(a\frac{d^2}{dx^2}\right)f(x)$ I know that
\begin{align*}
\exp\left(a\frac{d}{dx}\right)f(x)=f(x+a)\,,
\end{align*}
by comparing the Taylor expansions of both sides ($f(x)$ is an arbitrary function). However, if I have, where $f(x)$ is still arbitrary,
\begin{align*}
\exp\left(a\frac{d^2}{dx^2}\right)f(x)\,,
\end{align*}
is there also a simple form for this expression?
Expanding the exponential only results in the even derivatives of $f(x)$, so I am reasonably sure that this is not equal to something as simple as, for example,  $f(x+a)^2$. I then realized that
\begin{align*}
\exp\left(a\frac{d^2}{dx^2}\right)f(x)=\exp\left(a\frac{d}{dx}\right)^{\frac{d}{dx}}f(x)\,,
\end{align*}
If I expand the exponential, I get
\begin{align*}
\left(1+a\frac{d}{dx}+\frac{a^2}{2}\frac{d^2}{dx^2}+\cdots\right)^{\frac{d}{dx}}f(x)\,.
\end{align*}
However, now I am not sure how to treat this. I was thinking that this might require fractional calculus, but I am unsure how to proceed.
 A: Set $D=d/dx$. Whenever the series defining $\exp\left(a D^2 \right) f(x)$ converges, we have the identity
$$\exp\left(a D^2 \right) f(x) =\int_{-\infty}^\infty \frac{1}{\sqrt{4\pi a}}e^{-(x-y)^2/4a} \:f(y) \:dy$$
This integral transform is a scaled version of the Weierstrass transform.
Similarly we have 
$$\exp\left(a D^3 \right) f(x) =\int_{-\infty}^\infty \frac{1}{\sqrt[3]{3a}}\mathrm{Ai}(\frac{y-x}{\sqrt[3]{3a}}) \:f(y) \:dy$$
where $\mathrm{Ai}$ is the Airy function, and in general
$$\exp\left(a D^n \right) f(x) =\int_{-\infty}^\infty \mathcal{K}_n(x-y) \:f(y) \:dy$$
where the convolution kernel $\mathcal{K}_n(x)$ is the inverse Fourier transform (non-unitary, angular frequency) of $e^{a(-ip)^n}$, which can be expressed in terms of generalized hypergeometric functions ${}_0F_{n-2}$. This also works when $n$ is not a natural number. See this related question for how to find these expressions in the general case of $\exp(f(x, D))$, borrowing a "trick" from quantum mechanics.

Note that the functions $\mathcal{K}_n$ are solutions of the differential equation
$$an\frac{d^{n-1}\mathcal{K}_n(x)}{dx^{n-1}} + x\mathcal{K}_n(x) = 0$$
Indeed for $n=3$ this is a scaled Airy differential equation, for $n=2$ this gives a Gaussian, and even for $n=1$, the delta function $\delta(x+a)$ can be considered a "solution" of $xK_1(x)=-aK_1(x)$. This can be proven pretty straightforwardly from the expression of $\mathcal{K}_n$ as a Fourier transform.
