 High-energy shift from a central delta barrier

Put $c=m\kappa /{\hslash }^2$, $k_n=(n+1/2)\pi /a$, and ${\delta }_n=a({\lambda }_n-k_n)$. The quantization equation gives ${\delta }_n=\arctan (c/{\lambda }_n)$, so ${\lambda }_n{\delta }_n\to c$ and ${\delta }_n\to 0$. Therefore ${\lambda }_n^2-k_n^2\to 2c/a$, proving the displayed absolute energy shift. Its ratio to the unperturbed energy tends to zero. The normalized unperturbed even state has probability density $1/a$ at the origin, so its first-order delta-barrier expectation is also $\kappa /a$.