Find Normalizing constant

Question

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?

Answer 1

\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.

Answer 2

$\int_0^\infty e^{-\sqrt{x}} dx = 2$. Use the linear substitution $t ={x \over \theta^2}$ to get the scaled result.