Suppose that $p\geq 4$ is an even integer and $n$ is a sufficiently large integer. Define $$ a_q = \frac{1}{q!}\int_0^\infty \frac{\sin^{2n} t}{t^{p-q+1}}dt, \quad q=1,\dots,p-1. $$ I want to show that $\{a_q\}$ is decreasing for $q=1,\dots,p-1$.
I am not sure what a good way to deal with this is. I looked at $$ a_q - a_{q+1} = \frac{1}{q!}\int_0^\infty \frac{\sin^{2n}t}{t^{p-q+1}}\left(1-\frac{t}{q+1}\right)dt. $$ Splitting the integral on $[0,\infty)$ into $[m\pi,(m+1)\pi)$ for $m\geq 0$, and note that the integral on each interval $[m\pi,(m+1)\pi)$ is dominated by the integral over $[(m+\frac12-\delta)\pi,(m+\frac12+\delta)\pi)$ for a constant $\delta>0$. I think one can show that $$ \int_{(\frac12-\delta)\pi}^{(\frac12+\delta)\pi}\frac{\sin^{2n}t}{t^{p-q+1}}\left(1-\frac{t}{q+1}\right)dt > \left|\sum_{m=1}^\infty \int_{(m+\frac12-\delta)\pi}^{(m+\frac12+\delta)\pi}\frac{\sin^{2n}t}{t^{p-q+1}}\left(1-\frac{t}{q+1}\right)dt\right| $$ Then by lower-bounding the left-hand side and upper-bounding the right-hand side, the Hurwitz zeta function pops up and the whole thing boils down to an inequality about the the Hurwitz zeta function, which might be hard to prove.
Is there a better way to prove this?
Not an answer so far, but a promising attempt.
The Laplace transform of $\sin(x)^{2n}$ is given by $$ \frac{(2n)!}{s(4+s^2)(16+s^2)\cdot\ldots\cdot(4n^2+s^2)} $$ so we have $$ a_q = \frac{(2n)!}{q!(p-q)!}\int_{0}^{+\infty}\frac{s^{p-q-1}}{(4+s^2)(16+s^2)\cdot\ldots\cdot(4n^2+s^2)}\,ds$$ or $$ a_q = \frac{\frac{1}{4^n}\binom{2n}{n}}{q!(p-q)!}\int_{0}^{+\infty}\frac{s^{p-q-1}\,ds}{\left(1+\frac{s^2}{4}\right)\left(1+\frac{s^2}{16}\right)\cdot\ldots\cdot\left(1+\frac{s^2}{4n^2}\right)} $$ which seems more manageable, resembling $$ \widetilde{a_q} = \frac{\frac{1}{4^n}\binom{2n}{n}}{q!(p-q)!}\int_{0}^{+\infty}s^{p-q-1}e^{-\frac{\pi^2}{24}s^2}\,ds $$ with a reasonably concise closed form.