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Let $f(x,t) = -x^3 e^{-tx^3}$

I'm trying to find a dominating, integrable function over $f$ for all $t \in \mathbb{R}^+$. Specifically, I'm looking for a function $h$ s.t. $\forall t > 0$, we have $|f(x,t)| \le h(x), \forall x \ge 0$. My original idea was to try to show that $min\{\frac{1}{t^2x^5}, 1\}$ was such a function, but it failed in light of this post: Exponential Function Question.

I have already shown that

$$|-x^3 e^{-tx^3}| = |x^3||e^{-tx^3}| \le |-x^3||e^{-tx}|$$

so that if I could find an integrable dominator of $|-x^3||e^{-tx}|$ my original objective would follow. Any ideas?

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You say "for each fixed $t \in {\mathbb R}^+$", but that's not what you're really looking for: you want one that works for all $t > 0$. – Robert Israel Oct 30 '12 at 19:39
Thanks -- fixed in the original post. – user1770201 Oct 30 '12 at 19:44
up vote 2 down vote accepted

There is no such $h$. In fact, as $t \to 0+$ we have $f(x,t) \to -x^3$, so you'd need $h(x) \ge x^3$, and then $h$ is not integrable.

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What if I restricted the requirement from "for all t > 0" to "for all $t$ in some positive interval (a,b)"? Then we would avoid the problem of $t$ approaching $0$. Would this work? – user1770201 Oct 30 '12 at 19:46
Yes, if $a > 0$: just take $h(x) = x^3 e^{-ax^3}$. – Robert Israel Oct 30 '12 at 20:41
Thanks -- that works. Now I just need to show that such $h(x)$ is integrable. How would I do this? My starting idea would be to try to show that that $h(x) \le x^3 \cdot \frac{1}{a^2x^5} = \frac{1}{a^2x^2}$, the rightmost term known to be integrable from a prior exercise (so that since it dominates $h(x)$, it would follow $h(x)$ is also integrable). This inequality I think would be true if $e^{-ax^3} \le \frac{1}{a^2x^5}$, but this isn't in general true I don't think. – user1770201 Oct 31 '12 at 9:27
That is true for large enough $x$ : $e^{ax^3} \ge (a x^3)^2/2$ so $e^{-ax^3} \le \dfrac{2}{a^2 x^6} \le \dfrac{1}{a^2 x^5}$ if $x \ge 2$. – Robert Israel Oct 31 '12 at 18:04

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