I want to find the law of $B_T$, where $B_t$ is a brownian motion, $T$ is exp-distributed with parameter 1, with $B_t$ and $T$ being independent.
My idea is to say that the density of $B_T$ is given by $f_{B_T}(x)=\int_0^{\infty}f_{(B_T,T)}(x,y)\mathbb{d}y=\int_0^{\infty}f_{B_T|T}(x|y)*f_T(y)\mathbb{d}y$ where $f_{B_T|T}(x|y)=\frac{1}{\sqrt{2\pi y}}*e^{-\frac{x^2}{2y}}$ and $f_T(y)=e^{-y}$, since $B_T$ given $T=y$ is $\mathcal{N}(0,y)$-distributed.
Is this a proper argument?
Maybe more simply, and assuming that $T$ is independent of $(B_t)$, one has, for every bounded measurable function $u$ and every $t$, $$ E[u(B_T)\mid T=t]=E[u(B_t)\mid T=t]=E[u(B_t)]=\int_\mathbb R u(x)p_t(x)\mathrm dx, $$ where $(p_t)$ is the Brownian transition kernel. Integrating this with respect to the distribution of $T$ yields $$ E[u(B_T)]=E[E[u(B_T)\mid T]]=\int_0^\infty\left(\int_\mathbb R u(x)p_t(x)\mathrm dx\right)f_T(t)\mathrm dt, $$ where $f_T$ is the density of $T$. By Fubini, $B_T$ has density $g$ defined by $$ g(x)=\int_0^\infty p_t(x)f_T(t)\mathrm dt. $$