A quirky proof for a limit of an integral I am stuck on this result, which the professor wrote as "trivial", but I don't find a way out.
I have the function 
$$f_{\alpha}(t) = \frac{1}{2\pi} \sum_{k = 1}^{+\infty} \frac{1}{k}\int_0^{\pi} (\alpha(p))^k \sin^{2k}(\epsilon(p) t)\ dp$$
and he told use that for $t\to +\infty$ we have:
$$f_{\alpha}(t) = \frac{1}{2\pi} \sum_{k = 1}^{+\infty} \frac{1}{4^k k}\binom{2k}{k}\int_0^{\pi} (\alpha(p))^k\ dp$$
Now, it's all about the sine since it's the only term with a dependance on $t$. Yet I cannot find a way to send
$$\sin^{2k}(\epsilon(p) t)$$
into
$$\frac{1}{4^k}\binom{2k}{k}$$
Any help? Thank you so much.
More Details
$$\epsilon(p)$$
Is a positive, limited and continuous function.
The "true" starting point was
$$f_{\alpha}(t) = -\frac{1}{2\pi}\int_0^{\pi} \log\left(1 - \alpha(p)\sin^2(\epsilon(p)t)\right)\ dp$$
Then I thought I could have expanded the logarithm in series. Maybe I shouldn't had to...
 A: We can show this if we make an additional assumption: that $\epsilon(\cdot)$ is nowhere constant in the domain $[0\,\pmb,\,\pi]$. Let us consider the integral over a small part, say $a\leqslant p\leqslant a+\varepsilon$, of this domain. If $\varepsilon$ is small enough, then our function may be well approximated by $\epsilon(p)\approx bp+c$, where the constants $b$ and $c$ are respectively nonzero (by our additional assumption) and positive. Assuming also that $\alpha(\cdot)^k$ is continuous in our small interval, so that we can approximate it well by a constant, we can evaluate the integral $\int_a^{a+\varepsilon} \alpha(p)^k \sin^{2k}t\epsilon(p)\,\mathrm dp$ approximately as $$\int_a^{a+\varepsilon} \alpha(p)^k \sin^{2k}(tbp+tc)\,\mathrm dp.$$Using a change of variable to $x=tb(p-a)$, with $t=n\pi/\varepsilon b$, where $n$ is a very large natural number, we can write this as$$\frac{\varepsilon}{n\pi}\int_0^{n\pi} \alpha\left(\frac {\varepsilon x}{n\pi}+a\right)^k \sin^{2k}\left(x+\frac{n\pi(a+c)}{\varepsilon}\right)\,\mathrm dx.$$Note the small range of variation in the argument of $\alpha$. This allows us to treat the first factor of the integrand as near-constant, and write the integral approximately as $$\int_a^{a+\varepsilon}\alpha(p)^k\,\mathrm dp\cdot\frac1{n\pi}\int_0^{n\pi}\sin^{2k}\left(x+\frac{n\pi(a+c)}{\varepsilon}\right)\,\mathrm dx.$$By a standard result for the second integral (see note below), this reduces to $$\frac{1}{4^k}\binom{2k}{k}\int_a^{a+\varepsilon}\alpha(p)^k\,\mathrm dp.$$Now reassemble the original integral from the $\varepsilon$-length sections and let $n\to\infty$ and $\varepsilon\to0$. Our approximations become exact, and we end up with $$\frac{1}{4^k}\binom{2k}{k}\int_0^{\pi}\alpha(p)^k\,\mathrm dp.$$
If the $\epsilon$ function is constant over part of its domain, then this result is no longer valid, as remarked in the comments to the question.
Note: The integral $\int_0^{n\pi}\sin^{2k}(x+h)\,\mathrm dx$, where $k,n\in\Bbb N$ and $h$ is a constant, is independent of $h$; that is, $$\int_0^{n\pi}\sin^{2k}(x+h)\,\mathrm dx=\int_0^{n\pi}\sin^{2k}x\,\mathrm dx.$$ To see this, first change the variable by $u=x+h$, so that the left-hand integral may be written as $$\int_h^{n\pi+h}\sin^{2k}u\,\mathrm du.$$ By splitting the interval $[0\,\pmb,\,n\pi+h]$ at $h$ and at $n\pi$, we have$$\int_h^{n\pi+h}\sin^{2k}u\,\mathrm du=\int_0^{n\pi}\sin^{2k}u\,\mathrm du-\int_0^h\sin^{2k}u\,\mathrm du+\int_{n\pi}^{n\pi+h}\sin^{2k}u\,\mathrm du.$$Now we can rename the variable as $x$ in the first integral on the RHS, and see that the remaining two integrals cancel—say by a change of variable to $v=u-n\pi$ in the last integral.
It remains to show that$$\int_0^{n\pi}\sin^{2k}x\,\mathrm dx=n\int_0^\pi\sin^{2k}x\,\mathrm dx=\frac{(2k)!n\pi}{4^k(k!)^2}.$$We can simply look up $\int_0^\pi\sin^{2n}x\,\mathrm dx=(2n)!\pi/4^n(n!)^2$ in a table of standard integrals. Otherwise, the elementary way to prove this is by induction, splitting $\sin^{2k+2}x$ as $\sin^{2k+1}x\cdot\sin x$ and applying integration by parts.
A: Notice first that the series expansion makes sense only when $\alpha$ is bounded by $1$. This is because the radius of convergence of $\log(1+x)$ is $1$. In this answer, I will assume a slightly stronger condition on $\alpha$:

Assumption 1. $\sum_{k=1}^{\infty}\frac{1}{k} \int_{0}^{\pi}|\alpha(p)|^k \, dp < \infty$.

This assumption is indeed a stronger condition in that it implies $\|\alpha\|_{L^{\infty}} \leq 1$. This also allows us to freely interchange $t\to\infty$ limit and the infinite summation.
Now let me provide two different directions.

1st direction. As we have seen in comments, not all $\epsilon$ makes $f_{\alpha}(t)$ converge as $t\to\infty$ even when $\epsilon$ is assumed to be continuous. So we are led to impose some assumption, and the following turns out to be enough:

Assumption 2. $\epsilon$ is measurable and $\omega(y) = \operatorname{Leb}(\{ p \in [0, \pi] : \epsilon(p) \leq y \})$ is absolutely continuous.

Now define
$$ F_k(y) = \int_{0}^{\pi} \alpha(p)^k \mathbf{1}_{\{\epsilon(p) \leq y \}} \, dp. $$
Then we find that $|F_k(y') - F_k(y)| \leq \|\alpha\|_{L^{\infty}}^k |\omega(y') - \omega(y)|$ and thus $F_k$ is absolutely continuous. From this, it is not hard to check that
$$ \int_{0}^{\pi} \alpha(p)^k \sin^{2k}(\epsilon(p)t) \, dp
= \int_{-\infty}^{\infty} F_k'(y) \sin^{2k}(yt) \, dy. $$
(This is essentially 'conditioning on values of $\epsilon$'.) Then by expanding
\begin{align*}
\sin^{2k}(yt)
&= \left( \frac{e^{iyt} - e^{-iyt}}{2i} \right)^{2k} \\
&= \frac{1}{4^k} \sum_{j=0}^{2k} \binom{2k}{j} (-1)^{k-j}e^{(2k-2j)iyt} \\
&= \frac{1}{4^k}\binom{2k}{k} + \frac{2}{4^k} \sum_{j=0}^{k-1} \binom{2k}{j} (-1)^{k-j} \cos((2k-2j)yt)
\end{align*}
and applying the Riemann-Lebesgue lemma, we obtain
\begin{align*}
\lim_{t\to\infty} \int_{0}^{\pi} \alpha(p)^k \sin^{2k}(\epsilon(p)t) \, dp
&= \lim_{t\to\infty} \int_{-\infty}^{\infty} F_k'(y) \sin^{2k}(yt) \, dy \\
&= \frac{1}{4^k}\binom{2k}{k} \int_{-\infty}^{\infty} F_k'(y) \, dy \\
&= \frac{1}{4^k}\binom{2k}{k} \int_{0}^{\pi} \alpha(p)^k \, dp.
\end{align*}
Now by the dominated convergence theorem, we can pass the limit inside the sum and therefore
$$ \lim_{t\to\infty}f_{\alpha}(t)
= \frac{1}{2\pi}\sum_{k=1}^{\infty} \frac{1}{k \cdot 4^k}\binom{2k}{k} \int_{0}^{\pi} \alpha(p)^k \, dp. $$
Remark. Since the ultimate goal is to investigate the limiting behavior or $f_{\alpha}(t)$, I guess that we can weaken Assumption 1 and still obtain an analogous result:

Conjecture. If $\int_{0}^{\pi} \left|\log(1-\alpha(p))\right| \, dp < \infty$ and Assumption 2 holds, then $f_{\alpha}(t)$ converges as $t\to\infty$ and
$$\lim_{t\to\infty} f_{\alpha}(t) = -\frac{1}{\pi} \int_{0}^{\pi} \log \left( \frac{1 + \sqrt{1-\alpha(p)}}{2} \right) \, dp $$


2nd direction. Under Assumption 1, we can show that the Cesàro-mean (time-averaged mean) of $f_{\alpha}$ has the following limit:
$$ \lim_{T\to\infty} \frac{1}{T} \int_{0}^{T} f_{\alpha}(t) \, dt
= \frac{1}{2\pi}\sum_{k=1}^{\infty} \frac{1}{k \cdot 4^k}\binom{2k}{k} \int_{0}^{\pi} \alpha(p)^k \mathbf{1}_{\{\epsilon(p) \neq 0 \}} \, dp. $$
Indeed, by the dominated convergence theorem we can pass the time-averaged limit inside the summation to obtain
\begin{align*}
&\lim_{T\to\infty} \frac{1}{T} \int_{0}^{T} f_{\alpha}(t) \, dt \\
&= \frac{1}{2\pi}\sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{\pi} \alpha(p)^k \left( \lim_{T\to\infty} \frac{1}{T} \int_{0}^{T} \sin^{2k}(\epsilon(p)t) \, dt \right) \mathbf{1}_{\{\epsilon(p) \neq 0 \}} \, dp \\
&= \frac{1}{2\pi}\sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{\pi} \alpha(p)^k \left( \lim_{T\to\infty} \frac{1}{T|\epsilon(p)|} \int_{0}^{T|\epsilon(p)|} \sin^{2k}(t) \, dt \right) \mathbf{1}_{\{\epsilon(p) \neq 0 \}} \, dp
\end{align*}
Using the expansion of $\sin^{2k}(t)$ above, we can conclude that
$$ \lim_{T\to\infty} \frac{1}{T} \int_{0}^{T} \sin^{2k}(t) \, dt
= \frac{1}{4^k}\binom{2k}{k} $$
and hence the desired identity follows.
