Show that limit of integrals is zero Let $\mu$ be a Borel probability measure on $(0,+\infty)$ and $\alpha > 1$. Is it true that
$$
   \lim\limits_{y \to +0}  y^{\alpha-1} \int\limits_0^\infty x^2 e^{-y^{\alpha} x} \, d\mu(x) = 0\ ?
$$
If measure $\mu$ is concentrated in finite number of points or have a compact support it is true. But is it possible to say that it is true in general or there is a counterexample?
 A: If you choose $c>1$ and set
$$
d\mu(x) = \begin{cases}
\frac{c-1}{x^c}\,dx, &\text{if } x>1, \\\
0, &\text{if } x\le1,
\end{cases}
$$
then I believe the limit equals $+\infty$ for all $a>1$ and $c>1$.
A: This is not always true: if $\mathrm d\mu(x)=x^{-2}\mathbf 1_{x\geqslant1}\mathrm dx$, the LHS is $y^{-1}\mathrm e^{-y^\alpha}\to+\infty$ when $y\to0^+$.
To get a positive result, note that $y^{\alpha-1}x^{1-1/\alpha}\mathrm e^{-y^\alpha x}\leqslant c_\alpha$ uniformly on every nonnegative $x$ and $y$, for some finite $c_\alpha$. Hence the LHS is at most $c_\alpha\mathbb E(X^{1+1/\alpha})$, where the distribution of $X$ is $\mu$. Thus, if $\mathbb E(X^{1+1/\alpha})$ is finite, the limit of the LHS when $y\to0^+$ is zero. In fact, studying the case of the measures $\mathrm d\mu(x)=\beta x^{-\beta-1}\mathbf 1_{x\geqslant1}\mathrm dx$ with $\beta$ positive, one can guess that, if $\mathbb E(X^{\beta})$ is infinite for some $\beta\lt1+1/\alpha$, the limit of the LHS when $y\to0^+$ is not zero (our first counterexample being the case $\beta=1$).
