Prove, using the same method as in Proposition 6.2.5.1 that [begin{aligned}mathbb{P}_{r}^{(u)}left[frac{1}{R_{t}^{alpha}} exp left(-frac{mu^{2}}{2} int_{0}^{t} frac{d s}{R_{s}^{2}} ight)

Prove, using the same method as in Proposition 6.2.5.1 that

\[\begin{aligned}\mathbb{P}_{r}^{(u)}\left[\frac{1}{R_{t}^{\alpha}} \exp \left(-\frac{\mu^{2}}{2} \int_{0}^{t} \frac{d s}{R_{s}^{2}}\right)\right] & =\mathbb{P}_{r}^{(0)}\left[\frac{R_{t}^{u}}{r^{u} R_{t}^{\alpha}} \exp \left(-\frac{\mu^{2}+u^{2}}{2} \int_{0}^{t} \frac{d s}{R_{s}^{2}}\right)\right] \\& =\mathbb{P}_{r}^{(\gamma)}\left[\frac{R_{t}^{u-\gamma-\alpha}}{r^{u-\gamma}}\right],\end{aligned}\]

and compute the last quantity.

Proposition 6.2.5.1:

The joint Laplace transform of the pair \(\left(R_{t}^{2}, \int_{0}^{t} \frac{d s}{R_{s}^{2}}\right)\) satisfies

\[\begin{align*}
\mathbb{P}_{r}^{(u)} & \left\{\exp \left(-a R_{t}^{2}-\frac{\mu^{2}}{2} \int_{0}^{t} \frac{d s}{R_{s}^{2}}\right)\right\}=\mathbb{P}_{r}^{(\gamma)}\left\{\left(\frac{R_{t}}{r}\right)^{u-\gamma} \exp \left(-a R_{t}^{2}\right)\right\}  \tag{6.2.10}\\
& =\frac{r^{\gamma-u}}{\Gamma(\alpha)} \int_{0}^{\infty} d v v^{\alpha-1}(1+2(v+a) t)^{-(1+\gamma)} \exp \left(-\frac{r^{2}(v+a)}{1+2(v+a) t}\right)
\end{align*}\]

where \(\gamma=\sqrt{\mu^{2}+u^{2}}\) and \(\alpha=\frac{1}{2}(\gamma-u)=\frac{1}{2}\left(\sqrt{\mu^{2}+u^{2}}-u\right)\).