Is $J(y)= \int_{[1, \infty)} \exp(\frac{-1}{y^2}(x-\frac{1}{y})^2) \, dx$ uniformly convergent on $(0,1)$? Is $$J(y)= \int_{[1, \infty)} \exp \left(\frac{-1}{y^2} \left(x-\frac{1}{y} \right)^2 \right)dx$$ uniformly convergent for $y \in (0,1)$? From what i can see, the Weierstrass M test does not work ...
 A: The answer is yes. It can be shown that 
$$J(y)={1\over 2}\sqrt{\pi}y\left(1+\Phi({1-y\over y^2})\right)$$
where $\Phi(z)={2\over\sqrt{\pi}}\int_0^ze^{-t^2}\,dt$.
Since $\Phi$ is continuous and $\lim_{y\to 0}\Phi({1-y\over y^2})=2$, we see that $J(y)$ is continuous on $(0,1)$ and can be extended to a continuous function $\tilde{J}(y)$on $[0,1]$ (because the limit as $y\to 0$ exists and is equal to zero). The result follows because $\tilde{J}(y)$ is uniformly continuous on $[0,1]$.
A: Try plotting $\exp \left(\frac{-1}{y^2} \left(x-\frac{1}{y} \right)^2 \right)$ as $y \to 0$ and notice that not only it converges to $0$, but mass is also being lost on its way to infinity (imagine the integrand as being a sort of density of a Gaussian random variable with mean $\mu=\frac{1}{y}$ and variance $\sigma^2=\frac{y^2}{2}$ with the normalizing factor $1/\sigma\sqrt{2\pi}$ missing).
That being said, for a fixed $\epsilon$, you can choose a $\delta$ such that $\big|J(y)\big|<\epsilon$ for $y \in (0,\delta)$ and a large enough $M$ such that $\bigg|\int_{[M, \infty)} \exp \left(\frac{-1}{y^2} \left(x-\frac{1}{y} \right)^2 \right)dx\bigg|<\epsilon$ for $y \in [\delta,1)$ since $y \in [\delta,1)$ implies that the 'means of the densities' are bounded $(\mu \in (1,1/\delta])$ and there is a compact where nearly all the mass is concentrated.
A: Let's call the integrand $f(x,y).$ Let $\epsilon>0.$ If $0<y<\epsilon,$ then
$$\tag 1 \int_{[1, \infty)} f(x,y)\,dx \le \int_{\mathbb R} f(x,y)\,dx = \int_{{\mathbb R}} e^{-x^2/y^2}\,dx =  y\int_{{\mathbb R}} e^{-x^2}\,dx < \epsilon \sqrt {\pi}.$$
Now for $y \in [\epsilon,1]$ and $x\ge 1/\epsilon^2,$
$$\tag 2\frac{1}{y^2}(x-1/y^2)^2 \ge 1\cdot (x-1/y^2)^2 \ge (x-1/\epsilon^2)^2.$$
Choose any $M > 1/\epsilon^2$ such that $\int_M^\infty e^{-(x-1/\epsilon^2)^2}\, dx < \epsilon.$ Then $(2)$ shows$\int_M^\infty f(x,y)\,dx < \epsilon,$ for all $y \in [\epsilon,1].$ Put this together with $(1)$ and we have the desired result.
