*Repost of [this](https://math.stackexchange.com/questions/3572784/polar-decomposition-of-the-volterra-integral-operator) Math.SE question due to a lack of answers (No one was able to help me find the closed form of $U_T$ and $|T|$ after two bounties). I also searched extensively online but couldn't find anything. A reference for the solution suffices.*

>Define the Volterra integral operator (or numerical differentiation operator)
$$
T: L^2([0,1]) \to L^2([0,1]), \ 
f(x) \mapsto \int_{0}^{x} f(y) dy.
$$
Find its polar decomposition $T = U_T | T |$.

First I tried to find $|T| := (T^* T)^{\frac{1}{2}}$.
We have
$$
T^*: L^2([0,1]) \to L^2([0,1]), \ 
f(x) \mapsto \int_{x}^{1} f(y) dy.
$$
and therefore
$$
T^* T: L^2([0,1]) \to L^2([0,1]), \ 
f(x) \mapsto \int_{x}^{1} \int_{0}^{y} f(z) dz dy.
$$
Now I got stuck when finding $\sqrt{T^* T}$. I found out that $T^2 \ne T^* T$ and $(T^*)^2 \ne T^* T$.
Can somebody please give me a hint?

I have also tried writing out $T^* T$ more explicitly: denote by $F$ the anti-derivative of $f$ and by $\mathscr{F}$ the antiderivative of $F$.
The we have
$$
(T^* T f)(x)
= \int_{x}^{1} F(y) - F(0) dy
= \int_{x}^{1} F(y) dy - (1 - x) F(0)
= \mathscr{F}(1) - \mathscr{F}(x) - (1 - x)F(0).
$$

**Edit.**
I also know that $T^* T$ is compact and self-adjoint and  
$$
(T^{n + 1} f)
= \frac{1}{n!} \int_0^x (x - y)^n f(y) dy.
$$
holds.
Maybe this can help use find a closed form for $(I - T^* T)^n$?

**Edit 2.**
I know that
$$
| T |
= \sum_{n \in \mathbb N} \sigma_n \langle \cdot, v_n \rangle v_n,
$$
where $\sigma_n := \frac{1}{\pi\left(n - \frac{1}{2}\right)}$ are the singular values of $T^* T$ and $(v_n)_{n \in \mathbb N} \subset L^2([0,1])$ a orthonormal basis of $\overline{\text{ran}(T)}$ which fulfils
$$
T^* T x
= \sum_{n \in \mathbb N} \sigma_n^2 \langle x, v_n \rangle, v_n.
$$
The eigenfunctions of $T^* T$ are $f_n(x) = C \cdot \cos(\sigma_n^{-1} x)$ for some $C \in \mathbb{R}$.

**Partial answer from Math.SE**
We have $$\sqrt{T^* T} = I-\sum^{\infty}_{n=1}a_n(I-T^*T)^n,$$ where the $a_n$ are determined by $$1-\sqrt{1-x}=\sum^{\infty}_{n=1}a_nx^n.$$
This yields $\sum_{n = 0}^{\infty} a_n = 1$. The first few $a_n$ are $$\left(\frac{1}{2}, 0, \frac{1}{8}, 0, \frac{1}{16}, 0, \frac{5}{128}, 0, \frac{7}{256}, 0, \frac{21}{1024}, 0, \frac{33}{2048}, 0, \frac{429}{32768}\right)$$