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.
Uniform convergence of $\int_0^\infty e^{-(x-\alpha)^2}dx$
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uniform-convergence
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1What precisely do you mean by "uniform convergence" of $I(\alpha)$? If you mean that $\int_0^N e^{-(x-\alpha)^2}\; dx \to I(\alpha)$ as $N \to \infty$ uniformly in $\alpha$, that is not true. – 2017-02-27
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0@RobertIsrael, improper integral $I(a) = \int_0^{+\infty}f(x, a), x \in [a, +\infty] $ uniformly convergent the parameter $y \in [c, d]$, if it convergent on this segment and if : $\forall \epsilon > 0 \: \exists A(\epsilon) \ge a : \forall R > A, \forall y \in [c, d] \Rightarrow \Big| \int_R^{+\infty}f(x, a)dx \Big| < \epsilon$ – 2017-02-27
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0More generally, integrals $$J(a)=\int_0^\infty f(x-a)dx$$ never converge uniformly for $a$ in a set unbounded above (except if $f=0$). – 2017-02-27
1 Answers
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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 $\epsilon < \sqrt{\pi}/2.$
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0@marka_17: Contrary to some answers, the integral is not uniformly convergent. – 2017-02-27