We denote for $x \in \mathbb{R}_{+}$ and $m \in \mathbb{R}$ $$T_{m}(x) = \int_{0}^{\infty} t^{m} e^{-\left(t^{2}+x/t\right)} dt$$ For $t, x \in \mathbb{R}_{+}^{*}$, we set $g(t,x) = t^{2} + x/t$.
a) Show that for $x \in \mathbb{R}_{+}^{*}$, the function $t \in \mathbb{R}_{+}^{*} \mapsto g(t,x)$ is convex and admits a unique minimum at $t = M(x)$, which we will determine. Calculate $g(M(x),x)$.
b) Let $m \in \mathbb{R}_{+}$, and $x \in \mathbb{R}_{+}^{*}$. Show using the inequality $M(x) \leq x^{1/3}$ that $$T_{m}(x) \leq \int_{0}^{x^{1/3}} t^{m} e^{-g(t,x)} dt + e^{-\frac{19}{10} x^{2/3}} \left(\int_{x^{1/3}}^{\infty} t^{m} e^{-\frac{t^{2}}{20}} dt\right)$$
c) Show that for every $\varepsilon > 0$ (and $m \in \mathbb{R}_{+}$), we can find $C > 0$ such that $$\forall t \geq 1, \quad t^{m} \leq C t e^{\varepsilon t^{2}}$$ Deduce that for every $\varepsilon > 0$, $$e^{-\frac{19}{10} x^{2/3}} \left(\int_{x^{1/3}}^{\infty} t^{m} e^{-\frac{t^{2}}{20}} dt\right) = O_{x \rightarrow \infty}\left(e^{-\left[\frac{39}{20} - \varepsilon\right] x^{2/3}}\right)$$
d) Show that $$T_{m}(x) = O_{x \rightarrow \infty}\left(x^{\frac{m+1}{3}} e^{-3(x/2)^{2/3}}\right)$$