Find Normalizing constant let $f(x,\theta)=C_\theta \exp(-\sqrt{x}/\theta)$ where $x$ and $\theta$ are both positive.
Find the normalising constant $C_\theta$. I get $C_\theta=\sqrt{2}/\theta$ but my book says $C_\theta=1/2\theta^2$. Who is right?

\begin{align}
u & = \sqrt{x}/\theta \\
u^2 & = x/\theta^2 \\
2u\,du & = dx/\theta^2
\end{align}
$$
\int_0^\infty e^{-\sqrt{x}/\theta} \, dx = \theta^2\int_0^\infty e^{-u} \Big( 2u\,du \Big) = 2\theta^2.
$$
The integral can be done by parts, thus:
$$
\int u \Big(e^{-u}\,du\Big) = \int u\,dv=uv-\int v\,du = -ue^{-u}-\int -e^{-u}\,du,\text{ etc.}
$$
Then there's the problem of finding $-ue^{-u}\Big\vert_{u=0}^{u\to\infty}$.  That can be done by L'Hopital's rule.
