We denote for $x \in \mathbb{R}_{+}$ and $m \in \mathbb{R}$ $$T_{m}(x) = \int_{0}^{\infty} t^{m} e^{-\left(t^{2}+x/t\right)} dt$$
Let $L > 0$, $\rho \in C([0,L])$, and $g_{0}$ continuous and integrable on $\mathbb{R}_{+}$. We set for $x \in [0,L]$ and $v \in \mathbb{R}^{*}$: $$\begin{gathered} g(x,v) = \frac{2}{\sqrt{\pi}} \frac{e^{-v^{2}}}{|v|} \int_{x}^{L} \rho(y) e^{-\frac{x-y}{v}} dy, \quad \text{if} \quad v < 0, \\ g(x,v) = \frac{2}{\sqrt{\pi}} \frac{e^{-v^{2}}}{|v|} \int_{0}^{x} \rho(y) e^{-\frac{x-y}{v}} dy + g_{0}(v) e^{-\frac{x}{v}}, \quad \text{if} \quad v > 0. \end{gathered}$$
a) Show that $\alpha : x \in [0,L] \mapsto \int_{0}^{\infty} g_{0}(v) e^{-\frac{x}{v}} dv$ defines a function in $C([0,L])$.
b) Show that for $v \in \mathbb{R}^{*}$, the function $x \in [0,L] \mapsto g(x,v)$ is of class $C^{1}$ on $[0,L]$ and $$\begin{aligned} & \forall x \in [0,L], v \in \mathbb{R}^{*}, \quad v \frac{\partial g}{\partial x}(x,v) = \rho(x) \frac{2}{\sqrt{\pi}} e^{-v^{2}} - g(x,v) \\ & \forall v \in \mathbb{R}_{+}^{*}, \quad g(0,v) = g_{0}(v), \quad \forall v \in \mathbb{R}_{-}^{*}, \quad g(L,v) = 0 \end{aligned}$$