hope this question is not too silly. Does the following inequality hold??
\begin{equation} \int_{l}^{w} \frac{t^{a-1}dt}{(1+ut)^{p+1}} \le \int_{0}^{\infty} \frac{t^{a-1}dt}{(1+ut)^{p+1}} = \frac{1}{u^a}B(a,p+1-a) \end{equation}
where $0 < l < w < \infty$ and $0<a$, $1<p$.
Common sense says yes...because of the integration limits ... $[0,\infty)$ sweeps a greater area... but in mathematics, common sense is usually misleading...thanks a lot.
Yes, since the integrand is real, and strictly positive almost everywhere, the integral is not increased (indeed, decreased) by shrinking the interval of integration.
If you want to get really technical, take the integrand as a measure $d\mu$ and use that $A \subset B \implies \mu(A) \leq \mu(B)$ for $\mu$-measurable sets $A$ and $B$. Intervals are $\mu$-measurable since they are Borel sets and the measure is absolutely continuous with respect to Lebesgue measure, etc...