Uniform convergence of integral with parameter $\int_0^\infty x^a \cdot e^{-x^{a}}d x$ Does the integral
$$\int_0^\infty x^a \cdot e^{-x^{a}}d x$$
with parameter $a\in (0,\infty)$ converge uniformly?
It is clear how to solve this problem if $ 0 < a < \infty$
But what about my case?
 A: I think $\displaystyle \dfrac{-1}{\log a}\left(\dfrac{d}{dx}\right)e^{-x^a}=x^ae^{x^a}$ and, therefore, $$\int_0^\infty x^ae^{-x^a}=\left(\lim_{n\rightarrow\infty}\dfrac{-1}{\log a}e^{-x^a}\right)\Big|_{x=0}^{n}=\dfrac{1}{\log a}$$
since, $\displaystyle \lim_{n\rightarrow\infty} -n^a\le \lim_{n\rightarrow\infty}-\log n=-\infty$ $\Box$

Edited : The above was wrong, I hope I edited into the right one. The integral is equivalent to, by substituting $x^a=u$, $\displaystyle \nu=\dfrac{1}{a}\int_{0}^\infty u^{1/a}e^{-u}du$ and by integration by parts and induction on $k$, $\displaystyle \nu = \dfrac{1}{a^k}\int_{0}^\infty e^{-u}\cdot u^{1/a-(k-1)} du$ we choose $k$ so that $0\le \dfrac{1}{a}-(k-1)<1$ and hence, $\displaystyle e^{-u}u^{1/a-(k-1)}\le e^{-u/2}$ for any $u\ge 1$ and consider that $\displaystyle 0\le\int_{0}^1e^{-u}u^{1/a-(k-1)}du\le\int_{0}^1u^{1/a-(k-1)}du=\dfrac{1}{2-k+1/a}$ . Consequently, $$0\le\nu=\int_{0}^\infty x^ae^{-x^a}dx\le\dfrac{1}{2-k+1/a}+\int_{1}^\infty e^{-u/2}du=c$$
for some $0\le c<\infty$ and the proof is complete $\Box$
A: For $\displaystyle\int_0^\infty f(x,a) \, dx$ to converge uniformly for $a \in (0,\infty)$ it is necessary that
$$\lim_{c \to \infty}\sup_{a \in (0,\infty)}\left|\int_c^\infty f(x,a) \, dx\right|= 0$$
Here, we have
$$\int_0^\infty x^a e^{-x^a} \, dx = \frac{1}{a} \int_0^\infty x^{1/a}e^{-x} \, dx,$$
and for all $c > 1$
$$\sup_{a \in (0,\infty)}\left| \frac{1}{a}\int_c^\infty x^{1/a} e^{-x} \, dx\right|> \sup_{a \in (0,\infty)}\frac{c^{1/a}}{a}\int_c^\infty e^{-x} \, dx = \sup_{a \in (0,\infty)}\frac{c^{1/a}e^{-c}}{a}= \sup_{a \in (0,\infty)}\frac{1}{a}e^{\left(\frac{\log c}{a} - c\right)}$$
Taking $a_c = \frac{\log c }{c}\in (0,\infty)$ we get
$$\sup_{a \in (0,\infty)}\left| \frac{1}{a}\int_c^\infty x^{1/a_c} e^{-x} \, dx\right|> \frac{1}{a_c}e^{\left(\frac{\log c}{a_c} - c\right)}= \frac{c}{\log c},$$
implying
$$\lim_{c \to \infty}\sup_{a \in (0,\infty)}\left| \frac{1}{a}\int_c^\infty x^{1/a} e^{-x} \, dx\right|= +\infty$$
Thus, the convergence is not uniform for $a \in (0,\infty)$.
