Let $0<\alpha<1$, $h > 0$ and consider the following integral $$\int_1^\infty r^{-\alpha } e^{-hr} dr$$ Is this integral bounded by a constant $C>0$ independent of $h >0$? i.e $$\int_1^\infty r^{-\alpha } e^{-hr} dr \leq C ?$$
Since $\alpha <1$ the natural inequality $$\int_1^\infty r^{-\alpha } e^{-hr} dr \leq \int_1^\infty r^{-\alpha } dr= \infty $$ doesn't lead to anything. Is there some other way (since anyway we are dealing with an exponential even if its exponent can be arbitrary small) or the statement is not true?
The statement is not true.
Let $F(\alpha,h)$ be given by
$$\begin{align} F(\alpha,h)&=\int_1^\infty r^{-\alpha}e^{-hr}\,dr\\\\ &=\frac1{h^{1-\alpha}}\int_h^\infty x^{-\alpha}e^{-x}\,dx\tag1 \end{align}$$
It is easy to see that $F$ is unbounded as $h\to 0^+$ and hence, there is no number $C$ that uniformly bounds $F$ from above.