= Solution
The initial <wavefunction> has unit norm, since $\int_0^a(2/a)\sin^2(\pi x/a)\,dx=1$. On its nonzero half, $-\hbar^2\psi''/(2m)=E_*\psi$ with $E_*=\hbar^2\pi^2/(2ma^2)$. A small domain qualification is important: the <derivative> jumps at zero, so the state is not in the operator domain of the Dirichlet <Hamiltonian operator>. Its finite energy <expectation value> is nevertheless defined by the <quadratic form of a positive quantum Hamiltonian>. The <energy form of a half-well sine state> gives
$$
\langle E\rangle=\frac{\hbar^2}{2m}\int_{-a}^a|\psi'(x)|^2dx
=\frac{\hbar^2}{2m}\frac{2\pi^2}{a^3}\int_0^a\cos^2(\pi x/a)\,dx
=\boxed{\frac{\hbar^2\pi^2}{2ma^2}}.
$$
This is also the rigorous weak interpretation of the printed $\int\psi H\psi$. Indeed the distributional second <derivative> is $\psi''=-(\pi/a)^2\psi+\sqrt{2/a}(\pi/a)\delta_0$ inside the well. The <Dirac delta> term pairs to zero because $\psi(0)=0$. One should not instead conclude that this state is an <energy eigenstate> from its equation on just the positive half.
Its coefficients in the full-well <energy eigenstates> are
$$
a_n=\frac{\sqrt2}{a}\int_0^a\sin\frac{\pi x}{a}\sin\frac{n\pi(x+a)}{2a}\,dx.
$$
For even $n$, sine <orthogonality> gives $a_2=-1/\sqrt2$ and all other even coefficients zero. For $n=2p+1$, set $u=x/a$ and use $\sin[n\pi(u+1)/2]=(-1)^p\cos(n\pi u/2)$. Product-to-sum integration then gives
$$
\int_0^1\sin\pi u\cos\frac{n\pi u}{2}\,du=\frac{4}{\pi(4-n^2)},\qquad
\boxed{a_{2p+1}=\frac{4\sqrt2(-1)^p}{\pi[4-(2p+1)^2]}}.
$$
Since these coefficients are $O(n^{-2})$, $\sum_nE_n|a_n|^2$ converges, confirming that the state has finite energy. With $E_n=n^2E_1$, the spectral expression is
$$
\langle E\rangle=\frac12E_2+\frac{32E_1}{\pi^2}\sum_{p=0}^{\infty}\frac{(2p+1)^2}{[(2p+1)^2-4]^2}.
$$
The kinetic-energy <quadratic form> equals this spectral sum: expanding the weak <derivative> in the corresponding orthonormal cosine modes and using <Parseval's identity> gives $\int|\psi'|^2=\sum_nk_n^2|a_n|^2$. Equating it to the directly integrated value $4E_1$ proves
$$
\boxed{1=\frac12+\frac8{\pi^2}\sum_{p=0}^{\infty}\frac{(2p+1)^2}{[(2p+1)^2-4]^2}}.
$$
In contrast, $\sum_nE_n^2|a_n|^2$ diverges, consistently showing why $H\psi$ is not an ordinary square-integrable <vector> even though the energy expectation is finite.
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