Uniform continuity $I(x)=\int_0^{\infty}e^{-xt^2}dt$

Question

Consider the uniform continuity of the improper integral:

$$I(x)=\displaystyle \int \limits_0^{\infty}e^{-xt^2}dt$$

I have some ideas.

1. Consider $D_1=[a;\infty], \, \forall a>0$.

So, $\exists \, x_0 : 0 <x_0 \le x \Rightarrow e^{-xt^2}\le e^{-x_0t^2}$.

And, we have $I(x)=\displaystyle \int \limits_0^{\infty}e^{-xt^2}dt = \frac{1}{\sqrt{x_0}}\displaystyle \int \limits_0^{\infty}e^{-(\sqrt{x_0}t)^2}d(\sqrt{x_0}t)=\frac{\sqrt{\pi}}{2\sqrt{x_0}}$

According to Weierstrass standard, $I(x)$ is uniformly continuous on $D_1$.

2. Consider $D_2=[a;-\infty], \, \forall a\le0$.
3. Consider $D_3=(0;-\infty].$

I think in the $2^{nd} \, \text{and} \, 3^{rd}$ case, $I(x)$ isn't uniformly continuous. But, I don't know how to proof that.

Answer 1

$I(x)=\infty$ if $ x\leq 0$. For $x>0$ the integral is finite but it tends to $\infty$ as $x \to 0+$. This implies that $I$ is not uniformly continuous on $[0,\infty)$. It is uniformly continuous on $[a,\infty)$ for any $a>0$ becasue it is continuous and (by DCT) $I(x) \to 0$ as $x \to \infty$.