= Solution
The source prints bare differentials $\mathrm d$ in both <integrals>; the integration variable is $t$, so we write them as $dt$.<Discrete mathematics, continuous mathematics, and probability examination (2015) by University of Oxford>[Printed page 10, Question 9(c)(iii), both displayed differentials omit the variable]{ref}
The <normalization of a probability distribution> for $Y$ gives
$$
1=\int_0^\infty f_Y(t)\,dt
=\frac1{\sqrt\pi}\int_0^\infty t^{-1/2}e^{-t}\,dt.
$$
Therefore
$$
\boxed{\int_0^\infty t^{-1/2}e^{-t}\,dt=\sqrt\pi.}
$$
Next, the <standard normal distribution> has $\mathbb E[Z]=0$ and $\operatorname{var}(Z)=1$, so the second <moment> is $\mathbb E[Z^2]=1$. Since $Y=Z^2/2$, <linearity of expectation> yields $\mathbb E[Y]=1/2$. Expressing this <expectation> using the <probability density function> gives
$$
\frac12=\int_0^\infty t f_Y(t)\,dt
=\frac1{\sqrt\pi}\int_0^\infty t^{1/2}e^{-t}\,dt.
$$
Consequently
$$
\boxed{\int_0^\infty t^{1/2}e^{-t}\,dt=\frac{\sqrt\pi}{2}.}
$$
These evaluate the <gamma function> at $1/2$ and $3/2$, respectively, directly from a transformed <normal random variable>.
Back to article page
