# Sign of integral with increasing power

Maybe a trivial question. I think it is true, but I do not succeed in proving.

Suppose $h(x)$ is such that for all $a\geq 0$ $$\int_a^\infty h(x) dx\geq0$$ Prove for all $a\geq 0$ $$\int_a^\infty x h(x) dx\geq0$$

If needed, one may assume $\lim_{x\rightarrow\infty}h(x)=0$ and/or continuity of $h$.

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I think I solved it. $$\int_{x=a}^\infty x h(x) dx = \int_{x=a}^\infty \int_{y=0}^x \rm{d}\it y h(x) \rm{d}\it x = \int_{y=0}^\infty\int_{x=\min(y,a)}^\infty h(x) \rm{d} \it x \rm{d} \it y \geq 0$$ where we used that by assumption for all $y\geq 0$ $$\int_{x=\min(y,a)}^\infty h(x) \rm{d} \it x \geq 0$$ – Alexander Oct 25 '12 at 14:38

## 1 Answer

Introduce $H(s)=\displaystyle\int_s^\infty h(x) \mathrm dx$ for every $s$. Then $\displaystyle\int_a^\infty xh(x) \mathrm dx=aH(a)+\int_a^\infty H(s)\mathrm ds$ hence $\displaystyle\int_a^\infty xh(x) \mathrm dx\geqslant0$ for every $a$ as soon as $H(s)\geqslant0$ for every $s$.

More generally, $\displaystyle\int_a^\infty u(x)h(x) \mathrm dx=u(a)H(a)+\int_a^\infty u'(s)H(s)\mathrm ds\geqslant0$ for every nonnegative nondecreasing function $u$ as soon as $H(s)\geqslant0$ for every $s$.

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