Hey guys, I'm trying to prove that the following sequence of functions uniformly converges in $x \in [0, 1]$:
$$\begin{align} f_0 (x) \equiv 1, f_n (x) &= \sqrt{x f_{n-1}(x)} \\ \therefore f_n (x) &= x^{(1-\frac{1}{2^n})} \end{align}$$

It was pretty simple to show that this sequnce of functions approaches $f(x) = x$ in the given interval. To show the sequence uniformly converges, I wanted to prove the sequence is uniformly Cauchy. I wasn't sure if my following argument was correct...

Consider the following inequality:

\begin{align}
J(a)&=\int_{0}^{+\infty}\frac{\sin a x}{x}e^{-ax^{4}}{\rm d}x\\
&=\int_{0}^{+\infty}\color{blue}{x^{-1}}e^{-ax^{4}}\sin(ax){\rm d}x\\
&=\int_{0}^{+\infty}\left[\frac{1}{\Gamma(1)}\int_{0}^{+\infty}x^{1-1}e^{-ux}{\rm d}u\right]e^{-ax^{4}}\sin(ax){\rm d}x\\
&=\int_{0}^{+\infty}\left[\int_{0}^{+\infty}e^{-ux}{\rm d}u \right]e^{-ax^{4}}\sin(ax){\rm d}x\\
&=\int_{0}^{+\infty}\left[\int_{0}^{+\infty}e^{-ux-ax^{4}}{\rm d}u \right]\sin(ax){\rm d}x\\
&=\int_{0}^{+\infty}\left[\color{red}{\int_{0}^{+\infty}e^{-ux-ax^{4}}\sin(ax){\rm d}x}\right]{\rm d}u