Can $e^{ax}$ be said to be the eigenfunction of the operator $\frac{d^{(n)}}{dx}$? I'm gradually getting familiar with operators (as they are used in QM) and the terminology surrounding them, and I was wondering whether all the (to me) well-known operators have straight-forward, elementary functions, as seems to be the case with $\frac{d^{(n)}}{dx}$, because $$\frac{d^{(n)}}{dx}e^{ax}=a^n e^{ax}$$
and could one say that the spectrum of these eigenfunctions is degenerate, since $a$ can vary (I know "spectrum" is usually used for the set of eigenvalues, but it seems appropriate here)? Is this the correct interpretation? If so, what is the eigenfunctions and -values for the following differentialoperators (if you could point me in the direction of a resource that either collects them or - even better - shows how they are obtained, that would be very acceptable):


*

*$\int dx$

*$\nabla$ (grad)

*$\nabla \cdot $ (div)

*$\nabla^2$ (Laplace)

*If you have a cool one, please do just throw it in there!


Thanks!
 A: For each value of $a$, $$\frac{d^n}{dx^n} e^{ax} = a^n e^{ax} ,$$
and so $e^{ax}$ is an eigenfunction of the operator linear differential operator $\frac{d^n}{dx^n}$ of eigenvalue $a^n$. Conversely, the $a^n$-eigenspace of this operator (regarded as a map, e.g., from $C^{\infty}(\Bbb R)$ to itself) has dimension $n$, and it's not too hard to write down an explicit basis thereof. (NB that for $a = 0$, the eigenspace is qualitatively different from the general case, as the solutions of $\frac{d^n}{dx^n} f = 0$ are just the polynomials of degree $< n$.) With a little more work (mostly involving passing to the complex setting), we can show that the spectrum of this operator is $\Bbb R$ itself.
For most of the other operators you mention, the input and output objects are of different types, so without more structure it doesn't make any sense to talk about eigenfunctions; for example, the gradient maps functions to vector fields. A little more subtly, the operator $f \mapsto \int f \,dx$ maps functions to equivalence classes of functions (where $f \sim \hat{f}$ iff $\hat{f} - f$ is a constant).
The exception to this is the Laplacian $\Delta := \nabla^2$; in this case, the eigenvalue equation,
$$\Delta f = -\lambda f$$ is essentially the interesting and well-studied Helmholtz equation. The eigenvalues and eigenfunctions in this case depend on the (fixed, and often bounded) domain $\Omega$ of $f$, or, if you like, compact Riemannian manifold $(\Omega, g)$ with boundary. For general $\Omega$ the eigenfunctions are complicated, but when $\Omega$ is the $n$-sphere, the eigenvalues are the rather tractable spherical harmonics, which are important, for example, in the quantum mechanics of the hydrogen atom.
For any such domain $\Omega$, it's particularly interesting to restrict attention to eigenfunctions that vanish on the boundary (i.e., the functions $f$ that, in addition to the above equation, satisfy $f\vert_{\partial \Omega} = 0$). In this setting, we can ask whether the spectrum of eigenvalues for such eigenfunctions determine $\Omega$ (say, up to isometry), or a little more poetically, "Can One Hear the Shape of a Drum?" This question was posed by Mark Kac in a rightfully celebrated 1966 article by that title (though the origins of this circle of ideas are rather earlier), with emphasis on bounded domains $\Omega \subset \Bbb R^2$, and this question wasn't answered (in this case) until 1992, in the negative.
Another operator familiar from vector calculus that has the same domain and codomain is the curl operator, $\nabla \times\,\cdot\,$, on $\Bbb R^3$ (or any $3$-dimensional Riemannian manifold); its eigenvectors are the subject of this question.
A: There is no "the" eigenfunction. The set of all eigenfunctions form a eigenspace. If $T:V \to V$ is a linear operator, the collection of all $v \in V$ satisfying
$$ Tv = \lambda v
$$
for some scalar $\lambda$ is the eigenspace of $T$. It is not hard to see if $v$ is an eigenfunction of $T$, then so is $\mu v$ for any scalar $\mu$.
The operators gradient and divergence are not endomorphic. That is, they don't map spaces to themslves, so they don't have an eigenspace.
The eigenfunctions of $\nabla^2$ are given by the Helmholtz equation.
A: Yes, it's an eigenfunction but there are a lot. In particular, you need to attach some $n$th roots of unity since what you're really doing is solving an $n$th order linear ODE with constant coefficients, i.e., finding all the roots $r$ of $r^n - \lambda$ where
$$
\frac{d^{(n)}}{dx} y(x) = \lambda \cdot y(x).
$$
