Uniform convergence of $\int_0^\infty e^{-(x-\alpha)^2}dx$

Question

How to research on the uniform convergence the following integral $$I(\alpha) = \int_0^\infty e^{-(x-\alpha)^2}dx, a \in [0, +\infty]$$

Intuitively, I do believe that it uniformly converge.

Answer 1

The integral is not uniformly convergent for $\alpha \in[0, \infty)$.

Note that

$$\left|\int_R^\infty e^{-(x-\alpha)^2} \, dx\right| = \int_{R-\alpha}^\infty e^{-u^2} \, du. $$

For any $R > 0$, no matter how large, choose $\alpha_R \in [0, \infty)$ where $ \alpha_R =R$ and we have

$$\left|\int_R^\infty e^{-(x-\alpha_R)^2} \, dx\right| = \int_{0}^\infty e^{-u^2} \, du = \frac{\sqrt{\pi}}{2}. $$

Notice that your definition of uniform convergence is violated when $\epsilon < \sqrt{\pi}/2.$