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.
 A: 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.
A: While it tuns out quickly that there is no "simple formula" for these eigenvalues one is led to look at them numerically. They are indeed real and alternate in sign, the largest one being positive. The following figure shows the results for $n=30$:

