Linear algebra : eigenvalues of an integral operator on polynomials 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.
