Consider the linear transformation $$ T : \left\{ \begin{array}{ccc} \mathbb{R}_n[X] & \to & \mathbb{R}_n[X] \\ P & \mapsto & \int_0^1 (X + t)^n\,P(t)\,dt \end{array}\right. $$
where $\mathbb{R}_n[X]$ denotes the vector space of polynomials with real coefficients and degree $\leqslant n$.
NB: $T$ is self-adjoint with respect to the $L^2$-inner product $\langle P, Q \rangle = \int_0^1 P(t)\, Q(t) \,dt$.
The question is: what are the eigenvalues of $T$?
Edit: This question was originally posted by someone on the French maths forum les-mathematiques.net (here). I do not know if there is a reason to think that it is actually possible to find an explicit expression for the eigenvalues.
Using the adhoc convention $\frac{x}{0}=x$, Newton’s binomial formula yields
$$ T(X^j)=\sum_{i=0}^{n} \frac{\binom{n}{i}}{n+1+j-i}X^i \tag{1} $$
If we denote by ${\cal B}_n=(1,X,X^2, \ldots, X^{n})$ the canonical basis of ${\mathbb R}_n[X]$, the matrix $A$ of $T$ relatively to ${\cal B}_n$ is $A=(a_{ij})_{1\leq i,j \leq n+1}$ where
$$ a_{ij}=\frac{\binom{n}{i-1}}{n+1+j-i} \tag{2} $$
For example, when $n=4$ we have
$$ A=\left(\begin{matrix} \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} \\ \ & & & & \\ 1 & \frac{4}{5} & \frac{2}{3} & \frac{4}{7} & \frac{1}{2} \\ \ & & & & \\ 2 & \frac{3}{2} & \frac{6}{5} & 1 & \frac{6}{7} \\ \ & & & & \\ 2 & \frac{4}{3} & 1 & \frac{4}{5} & \frac{2}{3} \\ \ & & & & \\ 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \end{matrix}\right) $$
The characteristic polynomial of this $A$ is
$$\chi_A=X^5 - \frac{16}{5}X^4 - \frac{251}{450}X^3 + \frac{2977}{330750}X^2 - \frac{19}{1890000}X + \frac{1}{2778300000}$$
a polynomial whose Galois group over $\mathbb Q$ is ${\mathfrak S}_5$ and whose roots are therefore not expressible by radicals.
Thus I concur with Christopher A.Wong’s comment that it is unlikely that a simple closed-form solution exists.