TITLE: Asymptotic solution to the integral $\int_{-\pi/2}^{\pi/2} (\alpha + \sin x)^n \cos^2 x\,\mathrm{d}x$
QUESTION [6 upvotes]: Recently, I have posted a question on how to find a reduction formula for the trigonometric integral
$$\int (\alpha + \sin x)^n \cos^2 x\,\mathrm{d}x.$$
This problem seems to be tough, however. When trying to solve (as proposed in a comment on my previous question) the integral using the Binomial theorem, one obtains a sum that seems to be quite complicated.
Thus, there seems to be only a little chance for a reasonable reduction formula to exist. As a consequence, I became interested in solving the integral asymptotically. So my question is how to obtain a reasonable asymptotic estimate for the following definite integral, as $n \to \infty$:
$$\int_{-\pi/2}^{\pi/2} (\alpha + \sin x)^n \cos^2 x\,\mathrm{d}x.$$
I have tried to use the Laplace's method, but with my small knowledge and without any experience, I have not succeeded. Thus, I would be grateful for hints.

REPLY [7 votes]: I'll continue the work started in David H's answer.
It was shown there that
$$
I_\alpha(n) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\alpha + \sin x)^n \cos^2 x\,dx = \int_{\alpha -1}^{\alpha+1} u^n\sqrt{1-(u-\alpha)^2}\,du.
$$
We consider the cases $\alpha \geq 1$, $0 < \alpha < 1$, $\alpha = 0$, $-1 < \alpha < 0$, and $\alpha \leq -1$.
Case 1: $\alpha \geq 1$.  Begin by making the substitution $u = (\alpha+1)e^{-t}$ to get
$$
I_\alpha(n) = (\alpha+1)^{n+1} \int_0^{-\log\left(\frac{\alpha-1}{\alpha+1}\right)} e^{-nt} e^{-t} \sqrt{1-\left[(\alpha+1)e^{-t}-\alpha\right]^2}\,dt.
$$
Since
$$
\begin{align}
&e^{-t} \sqrt{1-\left[(\alpha+1)e^{-t}-\alpha\right]^2}\,dt \\
&= \sqrt{2+2\alpha}\, t^{1/2} \left[1-\frac{\alpha+6}{4}t - \frac{3\alpha^2-36\alpha-116}{96}t^2 - \frac{\alpha^3-6\alpha^2+36\alpha+88}{128}t^3 + O(t^4)\right]
\end{align}
$$
as $t \to 0^+$, we can apply Watson's Lemma to obtain the asymptotic expression
$$
\begin{align}
I_\alpha(n) &= \sqrt{\frac{\pi}{2}}\frac{(\alpha+1)^{n+3/2}}{n^{3/2}}\left[1 -\frac{3 (\alpha+6)}{8 n} - \frac{5 \left(3\alpha^2-36\alpha-116\right)}{128 n^2} \right. \\
&\qquad\qquad \left. - \frac{105 \left(\alpha^3-6\alpha^2+36\alpha+88\right)}{1024 n^3} + O\left(\frac{1}{n^{4}}\right)\right].
\end{align}
$$
Using more terms in the series expansion of the integrand will yield more terms of the asymptotic expansion of $I_\alpha(n)$.
Case 2: $0 < \alpha < 1$.  We split the integral into the two pieces
$$
I_\alpha(n) = \int_{0}^{\alpha+1} u^n\sqrt{1-(u-\alpha)^2}\,du + \int_{\alpha -1}^{0} u^n\sqrt{1-(u-\alpha)^2}\,du = J_1(n) + J_2(n).
$$
In the first we make the substitution $u = (\alpha+1)e^{-t}$ as before, and in the second we make the substitution $u=(\alpha-1)e^{-t}$.  The integrals become
$$
J_1(n) = (\alpha+1)^{n+1} \int_0^{\infty} e^{-nt} e^{-t} \sqrt{1-\left[(\alpha+1)e^{-t}-\alpha\right]^2}\,dt,
$$
$$
J_2(n) = (-1)^n (1-\alpha)^{n+1} \int_0^{\infty} e^{-nt} e^{-t} \sqrt{1-\left[(1-\alpha)e^{-t}+\alpha\right]^2}\,dt.
$$
By applying Watson's lemma to each of these individually we find that
$$
J_1(n) \sim \sqrt{\frac{\pi}{2}}\frac{(\alpha+1)^{n+3/2}}{n^{3/2}}
$$
and
$$
J_2(n) \sim (-1)^n \sqrt{\frac{\pi}{2}}\frac{(1-\alpha)^{n+3/2}}{n^{3/2}},
$$
so the terms coming from the first integral dominate those coming from the second.  Thus $I_\alpha(n)$ has the same asymptotic expansion as in Case 1.
Case 3: $\alpha = 0$.  In this case $J_2(n) = (-1)^n J_1(n)$, so that
$$
I_0(n) = \Bigl(1+(-1)^n\Bigr)\int_0^{\infty} e^{-nt} e^{-t} \sqrt{1-e^{-2t}}\,dt.
$$
Here
$$
e^{-t} \sqrt{1-e^{-2t}} = \sqrt{2}\,t^{1/2} \left[1-\frac{3}{2}t+\frac{29}{24}t^2-\frac{11}{16}t^3 + O(t^4)\right],
$$
so applying Watson's lemma yields
$$
I_0(n) = \sqrt{\frac{\pi}{2}} \frac{1+(-1)^n}{n^{3/2}} \left[1-\frac{9}{4 n}+\frac{145}{32 n^2}-\frac{1155}{128 n^3} + O\left(\frac{1}{n^4}\right)\right].
$$
Case 4: $-1 < \alpha < 0$.  This case is similar to Case 2.  The only difference is that the terms from $J_2(n)$ dominate.  We get
$$
\begin{align}
I_\alpha(n) &= (-1)^n \sqrt{\frac{\pi}{2}} \frac{(1-\alpha)^{n+3/2}}{n^{3/2}} \left[1+\frac{3(\alpha-6)}{8 n} - \frac{5\left(3\alpha^2+36\alpha-116\right)}{128 n^2} \right. \\
&\qquad\qquad \left. + \frac{105 \left(\alpha^3+6\alpha^2+36\alpha-88\right)}{1024 n^3} + O\left(\frac{1}{n^4}\right)\right].
\end{align}
$$
Case 5: $\alpha < -1$.  This case is similar to Case 1 (and its relation to Case 2).  If we made the substitution $u=(\alpha-1)e^{-t}$ in $I_\alpha(n)$ and applied Watson's lemma we would find that $I_\alpha(n)$ has the same asymptotic expansion as in Case 4.
In summary,

If $a > 0$ then
  $$
\begin{align}
I_\alpha(n) &= \sqrt{\frac{\pi}{2}}\frac{(\alpha+1)^{n+3/2}}{n^{3/2}}\left[1 -\frac{3 (\alpha+6)}{8 n} - \frac{5 \left(3\alpha^2-36\alpha-116\right)}{128 n^2} \right. \\
&\qquad\qquad \left. - \frac{105 \left(\alpha^3-6\alpha^2+36\alpha+88\right)}{1024 n^3} + O\left(\frac{1}{n^{4}}\right)\right]
\end{align}
$$
  as $n \to \infty$.
If $a = 0$ then
  $$
I_0(n) = \sqrt{\frac{\pi}{2}} \frac{1+(-1)^n}{n^{3/2}} \left[1-\frac{9}{4 n}+\frac{145}{32 n^2}-\frac{1155}{128 n^3} + O\left(\frac{1}{n^4}\right)\right]
$$
  as $n \to \infty$.
If $a < 0$ then
  $$
\begin{align}
I_\alpha(n) &= (-1)^n \sqrt{\frac{\pi}{2}} \frac{(1-\alpha)^{n+3/2}}{n^{3/2}} \left[1+\frac{3(\alpha-6)}{8 n} - \frac{5\left(3\alpha^2+36\alpha-116\right)}{128 n^2} \right. \\
&\qquad\qquad \left. + \frac{105 \left(\alpha^3+6\alpha^2+36\alpha-88\right)}{1024 n^3} + O\left(\frac{1}{n^4}\right)\right]
\end{align}
$$
  as $n \to \infty$.

Below are some plots of the three regimes for $5 \leq n \leq 40$.  The blue dots are values of $I_\alpha(n)$ obtained numerically and the red dots are the asymptotic approximations obtained above.  The first two terms of the asymptotics are used.

Figure 1: $\alpha = 1/10$.

Figure 2: $\alpha = 0$.

Figure 3: $\alpha = -1/10$.