Is the integral of a smooth function continuous?
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Consider e.g. $$f(a,x) = e^{-(x - 1/a)^2} \ \text{for}\ a \ne 0, \ f(0,x) = 0$$
Then $I(a) = \sqrt{\pi}$ for $a \ne 0$ but $I(0) = 0$.
Additional conditions that guarantee continuity will typically allow the use of the Lebesgue dominated convergence theorem.
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M.B.M.
Updated on February 06, 2020Comments
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M.B.M. almost 3 years
Suppose I have a function $f(a,x):\mathbb{R}^2\rightarrow\mathbb{R}$ that is smooth (i.e. infinitely differentiable) over its entire domain $\mathbb{R}^2$. Let $I(a)=\int_{-\infty}^{\infty}f(a,x)dx$. Suppose that we know that $|I(a)|<\infty$ for all $a\in \mathbb{R}$. Is $I(a)$ continuous? If not, are there additional conditions on $f(a,x)$ that can guarantee continuity of $I(a)$?