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$$\begin{align}f(x, t)&=\frac{1-e^{-xt^2}}{t^2}\\ F(x)&=\int^\infty_0f(x,t)\ \mathrm{dt}\end{align}$$

I need to show that $F$ is continuous on $\Bbb R^+$. $F$ is defined everywhere and $f$ is continuous with respect to $x$, and now I need some uniform, integrable upper bound $g(t)$ to apply the dominated convergence theorem.

Obviously $\frac{1}{t^2}$ works between $1$ and $\infty$, but it's not integrable around $0$. I'd like to find some integrable function that uniformly bounds $f(x, t)$ with $t$ between $0$ and $1$, and then define $g$ piecewise.

The limit of $f(x, t)$ as $t\to0$ is $x$, so $f$ is certainly bounded, but it's a bound that depends on $x$, which doesn't help me.

Jack M
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2 Answers2

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We use these inequalities:

For $t\geq 1$ we have $$\frac{|1-e^{-xt^2}|}{t^2}\leq\frac{1}{t^2} ,\ \forall x\geq0$$

and for $0<t\leq1$ and using $e^x\geq 1+x$ we have for all $a>0$ $$\frac{|1-e^{-xt^2}|}{t^2}\leq a,\ \forall\ 0\leq x\leq a$$

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There won't be a bound which will not depend on $x$, as $\lim_{x\to \infty}f(x,t)=\frac 1{t^2}$, but it's not a problem since continuity is a local property. We indeed have to check that if $x_0\geqslant 0$ is fixed, then $F$ is continuous at $x_0$. So we can use it on $[x_0-\delta,x_0+\delta]$ for some small $\delta$ and the mean value theorem can give an integrable bound.

Davide Giraudo
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