= Solution
Write $k=\pi/a$, $A=\sqrt{2/a}$ and $\Psi_0=Ae^{ikx}\sin kx$. Its derivative is
$$
\Psi_0'=Ak e^{ikx}(\cos kx+i\sin kx),\qquad |\Psi_0'|^2=A^2k^2.
$$
<Integration by parts> with the <Dirichlet boundary conditions> gives the <energy expectation>
$$
\langle H\rangle_0=\frac{\hbar^2}{2m}\int_0^a|\Psi_0'|^2\,dx=\boxed{\frac{\pi^2\hbar^2}{ma^2}=2E_1.}
$$
It is \b[constant in time], since the time-independent <Hamiltonian operator> has fixed <energy measurement> probabilities: $\langle H\rangle_t=\sum_nE_n|c_n|^2$.
For the <midpoint parity operator>, $R\Psi_0=-Ae^{-ikx}\sin kx$. Hence, again putting $u=kx$,
$$
\langle R\rangle_0=-\frac2\pi\int_0^\pi e^{-2iu}\sin^2u\,du=\boxed{\frac12.}
$$
Indeed the imaginary part integrates to zero by reflection and $\int_0^\pi\cos2u\sin^2u\,du=-\pi/4$. This <expectation value> is also \b[constant in time], because the bounded <self-adjoint operator> $R$ commutes with $H$ and hence with its <time-evolution operator>. Equivalently, $\langle R\rangle_t=\sum_n(-1)^{n+1}|c_n|^2$.
For the <momentum operator>, direct evaluation gives
$$
\begin{aligned}
\langle P\rangle_0&=-i\hbar A^2k\int_0^a\sin kx(\cos kx+i\sin kx)\,dx\\
&=\hbar A^2k\frac a2=\boxed{\frac{\pi\hbar}{a}.}
\end{aligned}
$$
The real sine-cosine integral vanishes. This <expectation value> is \b[not constant in time]: the <infinite square well> walls reverse motion. One can prove the change exactly without approximating the <wavefunction>. At
$$
t_* =\frac{\pi\hbar}{E_1}=\frac{2ma^2}{\pi\hbar},\qquad e^{-iE_nt_*/\hbar}=e^{-i\pi n^2}=(-1)^n,
$$
the <energy eigenstate> expansion and $R\psi_n=(-1)^{n+1}\psi_n$ give $\Psi(t_*)=-R\Psi_0$. Differentiation shows $PR=-RP$, so
$$
\boxed{\langle P\rangle_{t_*}=-\langle P\rangle_0=-\frac{\pi\hbar}{a}.}
$$
This <half revival in an infinite square well> is a concrete counterexample to momentum conservation. The sign from $-R$ is only a <global phase>.
\Image[/past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2017-box-reflection.png]
{title=The box state at $0$ and $t_*$: identical position <probability density>, opposite <probability current>}
{description=Analytic curves in dimensionless coordinates; arrows indicate the current direction. The integrated current gives $\langle P\rangle/m$, so the reversed currents have opposite mean momentum.}
The apparent interior identity $[H,P]=0$ cannot justify conservation here: the <momentum operator> does not preserve the <Dirichlet boundary conditions> of the <Hamiltonian operator>'s <operator domain>. For sufficiently regular states, retaining the boundary terms in <Ehrenfest theorem> instead gives the <boundary force in an infinite square well>,
$$
\frac{d\langle P\rangle}{dt}=\frac{\hbar^2}{2m}\left(|\partial_x\Psi(0,t)|^2-|\partial_x\Psi(a,t)|^2\right).
$$
In the strictly confined interval, $-i\hbar\partial_x$ with both endpoint values zero is symmetric but not itself a <self-adjoint operator>. The requested integral is the conventional mean momentum; it also equals the full-line momentum mean of the zero-extended state, as measured upon releasing the walls. This domain distinction explains why the formal interior <commutator> misses the wall force.
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