= Solution
{c}
For $I_1$, the <change of variables> $t=x^2/\theta$ gives
$$
\boxed{I_1=\int_0^\infty e^{-t}\,dt=1.}
$$
For $k\geq1$, use $d(e^{-x^2/\theta})=-(2x/\theta)e^{-x^2/\theta}\,dx$ and <integration by parts>:
$$
\begin{aligned}
I_{k+2}&=-\int_0^\infty x^{k+1}\,d(e^{-x^2/\theta})\\
&=-\left[x^{k+1}e^{-x^2/\theta}\right]_0^\infty+(k+1)\int_0^\infty x^ke^{-x^2/\theta}\,dx\\
&=\boxed{\frac{(k+1)\theta}{2}I_k.}
\end{aligned}
$$
The boundary term vanishes: $x^{k+1}\to0$ at zero, and the <exponential decay> dominates the power at infinity. These estimates also establish convergence of the <improper integrals> used above.
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