= Solution
For the <adjoint of an integration operator anchored at zero>, use the <inner product> $\langle f,g\rangle=\int f\bar g$. On the positive half-interval, reversing the triangular integration order gives
$$
\int_0^\pi\left(\int_0^tf(s)ds\right)\overline{g(t)}dt=\int_0^\pi f(s)\overline{\left(\int_s^\pi g(t)dt\right)}ds.
$$
On the negative half-interval the corresponding calculation gives
$$
\int_{-\pi}^0\left(-\int_t^0f(s)ds\right)\overline{g(t)}dt=\int_{-\pi}^0f(s)\overline{\left(-\int_{-\pi}^sg(t)dt\right)}ds.
$$
These exchanges are justified by the <Fubini theorem>, since on a finite interval $L^2$ functions belong to $L^1$ and the absolute double integral is at most $\|f\|_1\|g\|_1$. Therefore
$$
(T^*g)(s)=\begin{cases}\displaystyle\int_s^\pi g(t)dt,&s>0,\\\displaystyle-\int_{-\pi}^sg(t)dt,&s<0.\end{cases}
$$
Adding $Tg$ gives $\int_0^\pi g$ on the positive half and $-\int_{-\pi}^0g$ on the negative half, exactly the prescribed $Ug$. Values at zero do not matter in $L^2$. Consequently
$$
\boxed{T+T^*=U.}
$$
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