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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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    Thanks -- fixed in the original post.2012-10-30

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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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    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 $x \ge 2$.2012-10-31