= Solution
{c}
The <probability density function> must integrate to one, so
$$
1=\frac1{\sqrt\pi}\int_0^\infty y^{-1/2}e^{-y}\,dy,\qquad \boxed{\int_0^\infty t^{-1/2}e^{-t}\,dt=\sqrt\pi.}
$$
Also, <linearity of expectation> and the <mean> and <variance> of the <standard normal distribution> give
$$
\mathbb E[Y]=\frac12\mathbb E[Z^2]=\frac12\left(\operatorname{var}(Z)+(\mathbb E[Z])^2\right)=\frac12.
$$
Using the <probability density function> to compute this <expected value> gives
$$
\frac12=\frac1{\sqrt\pi}\int_0^\infty y^{1/2}e^{-y}\,dy,\qquad \boxed{\int_0^\infty t^{1/2}e^{-t}\,dt=\frac{\sqrt\pi}{2}.}
$$
These are the values $\Gamma(1/2)$ and $\Gamma(3/2)$ of the <gamma function>.
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