Now asked on MathOverflow, too.
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 $$ (T^{n + 1} f) = \frac{1}{n!} \int_0^x (x - y)^n f(y) dy. $$ 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 $C \in \mathbb{R}$.
Regarding the sequence $(a_n)$ in Matematleta's answer: The first $a_{2n}$ are $$\left(\frac{1}{2}, \frac{1}{8}, \frac{1}{16}, \frac{5}{128}, \frac{7}{256}, \frac{21}{1024}, \frac{33}{2048}, \frac{429}{32768}\right),$$ the odd terms are zero. The numerators, omitting the zeros, are A098597 on OEIS and the denominators (omitting zeros again) are given by $2^{\ell}$, $\ell \in \{1,3,4,7,8,10,11,15\}$, which is this in OEIS.
The reason I hope there exists an explicit answer is that this exercise is taken from this problem set from LMU Munich wherein it states "Important: the full solution here is to produce the explicit "closed" form of the operators $U_T$ and $|T|$. Giving the canonical decomposition of $|T|$ is not enough, show that in fact $|T|$ is an integral operator and $U_T$ is a unitary operator (with an explicit, "easy" form)."
How about this sketch:
Notice that $f:x\mapsto 1-(1-x)^{1/2}$ has a power series expansion defined by $1-(1-x)^{1/2}=\sum^{\infty}_{n=1}a_nx^n$ which converges for $|x|<1.$ It is easy to show by letting $x\to 1^-$ that $\sum^{\infty}_{n=1}a_n=1$. We will apply this below.
First, normalize $T^*T$ so its norm is $1$. Then define $S:=I-\sum^{\infty}_{n=1}a_n(I-T^*T)^n$. Since $\|I-T^*T\|\le1,\ S$ is norm convergent. Now check that $S\ge 0$, is self-adjoint, and satisfies $S^2=T^*T.$ This gives a formula for $|T|$.
Now, since $\||T|f\|^2=\|Tf\|^2$ it follows that $T$ and $|T|$ have the same nullspace. And $U(|T|f):=Tf$ is an isometry mapping the range of $|T|$ onto the range of $T.$ Now set $Uf:=0$ for all $f\in \text{Ran}|T|^{\perp}$ (OK since $\text{Ran}\ |T|^{\perp} = \text{Ker}\ |T|$ ), and check that $T=U|T|.$