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Here $h(x)$ attains its maximum one at $x=1$, so $a=1$. For $x\le 0$, put $t=-x\ge 0$. Then $x^{2}h(x)=e^{-1/2}{t}^2{e}^{-t/2}$. Its derivative is $e^{-1/2}{e}^{-t/2}t(2-t/2)$, so its maximum is at $t=4$, with value $16e^{-5/2}$.

For $0\le x\le 1$, $x^{2}h(x)=x^2{e}^{(x-1)/2}$ is increasing and has maximum one. For $x\ge 1$, $x^{2}h(x)=x^2{e}^{-(x-1)/2}$ has logarithmic derivative $2/x-1/2$, so its maximum is at $x=4$, with value $16e^{-3/2}>1$. Therefore the Laplace ratio-of-uniforms envelope has the finite boundsThese calculations establish finiteness and identify the tight axis-aligned bounds, rather than relying only on exponential tail decay.