= Solution
For $b>0$, the <exponential variational trial for a square-root half-line well> satisfies the wall <boundary condition> and has norm $\int_0^\infty x^2e^{-2bx}dx=1/(4b^3)$. <integration by parts> has no boundary term and gives
$$
\langle T\rangle
=\frac{\hbar^2}{2m}
\frac{\int_0^\infty(1-bx)^2e^{-2bx}dx}{1/(4b^3)}
=\frac{\hbar^2b^2}{2m}.
$$
The supplied <gamma integral> gives
$$
\langle V\rangle=\frac{15a}{16}\sqrt{\frac\pi2}\,b^{-1/2},\qquad
E(b)=\frac{\hbar^2b^2}{2m}
+\frac{15a}{16}\sqrt{\frac\pi2}\,b^{-1/2}.
$$
Setting $E'(b)=0$ yields the unique minimum,
$$
\boxed{b_*=\left(\frac{ma}{\hbar^2}\frac{15}{32}
\sqrt{\frac\pi2}\right)^{2/5}.}
$$
At the minimum $\langle V\rangle=4\langle T\rangle$, so the variational ground-state estimate is
$$
\boxed{E_{\rm var}=\frac{5\hbar^2b_*^2}{2m}
=1.876493\left(\frac{\hbar^2a^4}{2m}\right)^{1/5}.}
$$
The <energy> diverges as $b\to0$ or $b\to\infty$, confirming that the stationary point minimizes the trial <energy>.
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