Let $\mu$ be an exponentially decreasing Borel measure on $\mathbb{R}^n_{+}$, i.e. there exists $r>0$ such that $$ \int\limits_{\mathbb{R}^n_{+}} e^{r|x|} \, \mu(dx) < \infty. $$ I want to show that for any $N \in \mathbb{N}$ there exists a polynomial $P_{N}(y) \in \mathbb{R}[y]$ of degree at most $N$ such that $$ \int\limits_{\mathbb{R}^n_{+}} e^{-x \cdot y} \, \mu(dx) - P_{N}(y) = o(|y|^{N}), \;\;\; y \to 0, \; y \in \mathbb{R}^n_{+}. $$ My first question is if there is a name of such property? If I'm not mistaken it is denoted by $C^{\infty}(0,\mathbb{R}^n_{+})$.
My second question is how to prove this. I expanded $e^{- x \cdot y}$ in series and tried to show that $$ \int\limits_{\mathbb{R}^n_{+}} e^{-x \cdot y} \, \mu(dx) = \mu(\mathbb{R}^n_{+})+ \sum\limits_{k=1}^{\infty} \frac{(-1)^k}{k!} \int\limits_{\mathbb{R}^{n}_{+}} (x \cdot y)^k \, \mu(dx) \tag{1} $$ but without success. I don't know how to deal with expression $(x \cdot y)^n$ under the integral sign. When $n$ tends to infinity and $y \neq 0$ it takes arbitrarily large values and I can't estimate it. If this equality is shown then I can take $$ P_{N}(y) = \sum\limits_{k=0}^{N} \frac{(-1)^k}{k!} \int\limits_{\mathbb{R}^n_{+}} (x \cdot y)^k \, \mu(dx). $$ Then the question is why $$ \sum\limits_{k=N+1}^{\infty} \frac{(-1)^k}{k!} \int\limits_{\mathbb{R}^n_{+}} (x \cdot y)^k \, \mu(dx) = o (|y|^N), \;\;\; y \to 0 $$ but I think that the answer to the latter question will follow from the proof of $(1)$.
Writing by Cauchy-Schwarz inequality $$\left|\int_{\Bbb R^n_+}(x\cdot y)^k\mu(dx)\right|\leqslant \frac{|y|^k}{r^k}\int_{\Bbb R^n_+}e^{r|x|}\mu(dx),$$ the problem reduces to show that $$\sum_{k\geqslant N+1}\frac{|y|^k}{k!r^k}=o(y^N).$$ But $\sum_{k\geqslant N+1}\frac{|y|^k}{k!r^k}\leqslant |y|^{N+1}e^{r|y|}$, giving what we wanted.