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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.

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  • $\begingroup$ As a starting point you may omit the $\cos$ term and use as suggested the binomial theorem $$\int_{-\pi/2}^{\pi/2} (\alpha + \sin x)^n dx =\sum_{k=0}^n {n \choose k}\alpha^{n-k}\int_{-\pi/2}^{\pi/2}\sin(x)^k dx.$$ Obviously the integrals vanish for odd $k$ and for $k=2m$ you have $$\int_{-\pi/2}^{\pi/2}\sin(x)^{2m} dx = \sqrt{\pi} \frac{\Gamma\left(m+\frac{1}{2}\right)}{\Gamma\left(m+1\right)}$$ $\endgroup$ Commented Oct 28, 2013 at 11:00
  • $\begingroup$ According to my computations the integral is $$\pi\sum_{k\geq 0} \binom{n}{2k} \left\{\binom{2k}{k}(1/2)^{2k} - \binom{2k+2}{k+1}(1/2)^{2k+2}\right\} \alpha^{n-2k}$$ maybe that will help somehow. $\endgroup$ Commented Oct 28, 2013 at 11:21

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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.

enter image description here

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

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Figure 2: $\alpha = 0$.

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

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  • $\begingroup$ Thank you for the detailed answer! I was up to accept the David H's answer after doing some work on it... But as I see, you have done virtually all of the hard work instead of me. :) $\endgroup$
    – 042
    Commented Oct 28, 2013 at 17:23
  • $\begingroup$ You're welcome! I couldn't resist -- the formula obtained by David H was perfect for the method. $\endgroup$ Commented Oct 28, 2013 at 17:32
  • $\begingroup$ Awesome job,+1. I was pretty sure my formula was gonna be the best place to start, but I'm apparently too rusty to go from there. =p $\endgroup$
    – David H
    Commented Oct 28, 2013 at 17:44
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Let me know if any of this needs more details but there is a a fair bit of algebra and it will cloud the discussion if I include it all. By symmetry and trig identities: $$\int _{-\pi/2 }^{\pi/2 }\! \left( \alpha+\sin \left( x \right) \right) ^{n} \left( \cos \left( x \right) \right) ^{2}{dx}=\frac{1}{4} \int _{-\pi }^{\pi }\! \left( \alpha+\sin \left( x \right) \right) ^{ n}\cos \left( 2\,x \right) {dx}+\frac{1}{4}\int _{-\pi }^{\pi }\! \left( \alpha+\sin \left( x \right) \right) ^{n}{dx} \tag{1}$$ and by two rounds of binomial expansion: $$\left( \alpha+\sin \left( x \right) \right) ^{n}=\sum _{k=0}^{n} {\frac {1}{{2}^{ k}}}{n\choose k}{\alpha}^{n-k}{i}^{k}\sum _{j=0}^{k}{k\choose j} \left( -1 \right) ^{j}{{\rm e}^{ix(2j-k)}} $$ so $(1)$ is equivalent to: $$ \sum _{k=0}^{n} \mathfrak{R}\left( {\frac {{n\choose k}{\alpha}^{n- k}{i}^{k}\sum _{j=0}^{k} \left( {k\choose j} \left( -1 \right) ^{j} \int _{-\pi }^{\pi }\!{{\rm e}^{ix \left( 2\,j+2-k \right) }}{dx} \right) }{{2}^{k+2}}} \right) +\sum _{k=0}^{n} \frac{ 1}{{2}^{k+2}}\left({n\choose k}{\alpha}^{n-k}{i}^{k}\sum _{j=0}^{k} {k \choose j} \left( -1 \right) ^{j}\int _{-\pi }^{\pi }\!{{\rm e}^{ix \left( 2\,j-k \right) }}{dx} \right) \tag{2}$$ where the real part crops up because I am using only one exponential from the cosine. Then from the integral rep of the Kronecker delta symbol: $$\int _{-\pi }^{\pi }\!{{\rm e}^{ix \left( 2\,j-k \right) }}{dx}=2\pi \delta_{0,2j-k}\quad,\quad \int _{-\pi }^{\pi }\!{{\rm e}^{ix \left( 2\,j+2-k \right) }}{dx}=2\pi \delta_{0,2j+2-k}$$ it is evident that we sum only over even $k$ picking out one $j$ term from each inner sum. After a bit of algebra and converting binomial coefficients to factorials we obtain: $$\begin{aligned} \int _{-1/2\,\pi }^{1/2\,\pi }\! \left( \alpha+\sin \left( x \right) \right) ^{n} \left( \cos \left( x \right) \right) ^{2}{dx}&=\frac{\pi}{2^{n+1}}\sum _{k=0}^{\lfloor n/2 \rfloor }{ \frac {n!\, \left( 2\,\alpha \right) ^{n-2\,k}}{ \left( k+1 \right) \left( k! \right) ^{2} \left( n-2\,k \right) !}}\\ &=\frac{\pi \,{ \alpha}^{n}}{2} {\mbox{$_2$F$_1$}(-\frac{n}{2},-\frac{n}{2}+\frac{1}{2};\,2;\,\frac{1}{\alpha^2})} \tag{3} \end{aligned}$$ where $\mbox{$_2$F$_1$}$ is a hypergeometric function . Alternatively the result can also be given as a Jacobi polynomial but the order is integer for odd $n$ only: $$\frac{\pi \,{ \alpha}^{n}}{2} {\mbox{$_2$F$_1$}(-\frac{n}{2},-\frac{n}{2}+\frac{1}{2};\,2;\,\frac{1}{\alpha^2})}=\frac{{\alpha}^{n}\pi}{n+1} P_{\frac{1}{2}(n-1)}^{1,-3/2-n}\left(1-\frac{2}{\alpha^2}\right) \tag{4}$$ for which there is a recursion relation etc. In maple you can convert hypergeometric functions to Jacobi polynomials and vice versa and I expect mathematica does something similar. I appreciate this does not instantly give you the asymptotics but I was thinking you might consider it a "reduction" (wikipedia cites a "Darboux formula" for the asymptotics of Jacobi polynomials at large order).

You also have the following relation, if: $$I_{{n}} \left( \alpha \right) =\int _{-1/2\,\pi }^{1/2\,\pi }\! \left( \alpha+\sin \left( x \right) \right) ^{n} \left( \cos \left( x \right) \right) ^{2}{dx}$$ then: $${\frac {d}{d\alpha}}I_{{n}} \left( \alpha \right) =nI_{{n-1}} \left( \alpha \right)$$

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    $\begingroup$ The relationship ${\frac {d}{d\alpha}}I_{{n}} \left( \alpha \right) =nI_{{n-1}} \left( \alpha \right)$ is an example of what some people refer to as "Feynman's Trick". It's one of those methods that everyone should know yet practically no one does. $\endgroup$
    – David H
    Commented Oct 28, 2013 at 14:33
  • $\begingroup$ Thank you for your nice and very inspiring answer! I will definitely use some of those tricks later on... $\endgroup$
    – 042
    Commented Oct 28, 2013 at 17:32
  • $\begingroup$ No worries it was enjoyable, as it happens the Jacobi polynomial can be written with integer order for both even and odd $n$ by applying the final derivative relation to odd order in $(4)$ but I think you got this covered now. Thanks! $\endgroup$ Commented Oct 28, 2013 at 21:04
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Hint: Use the substitution $$u=\alpha +\sin{x},$$ $$du=\cos{x}\space dx,$$ $$\cos{x}=\sqrt{1-\sin^2{x}}=\sqrt{1-(u-\alpha)^2}.$$

The third equation is valid on intervals of $x$ where cosine is nonnegative, as is the case for $-\frac{\pi}{2}\leq x \leq \frac{\pi}{2}.$ Then,

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\alpha + \sin x)^n \cos^2 x\,\mathrm{d}x=\int_{\alpha -1}^{\alpha+1} u^n\sqrt{1-(u-\alpha)^2}du.$$

Applying a simple shift substitution, $v=u-\alpha$, the integral becomes

$$ \int_{\alpha -1}^{\alpha+1} u^n\sqrt{1-(u-\alpha)^2}du = \int_{-1}^{1} (\alpha+v)^n\sqrt{1-v^2}dv .$$

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Look for the maximum value of $f(x)=(\alpha+\sin x)^n\cos^2x$. For large values of $n$, it will be near $\pm\pi/2$.
Let $x=\pi/2-y$. $y$ is going to be small, and then use Taylor series.
Take logs of $f(x)$, and differentiate, I get $$\frac{n\cos x}{\alpha+\sin x}-\frac{2\sin x}{\cos x}=0$$
Rearrange, to get $$n\sin^2y=2\cos y(\alpha+\cos y)$$
Expand in Taylor series: the leading term on the LHS is $ny^2$, on the RHS is $2(\alpha+1)$.
So I expect the maximum to be near $y=n^{-1/2}\sqrt{2\alpha+2}$.
Calculate the second derivative of $\log f(x)$ to find the width of the spike.
I get $4\sqrt{\pi}(\alpha+1)^{n+3/2}n^{-3/2}/e$.

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I was too lazzy ! Use the binomal expansion and replace Cos[x]^2 by (1 - Sin[x]^2). You then have two almost identical integrals, the general term in the series being Sin[x]^m. The result of its integration between -Pi/2 and Pi/2 is given by

(1+(-1)^m Sqrt[Pi] Gamma[(1+m)/2]) /(m Gamma[m/2])

Is this of any interest ?

Continuing the work, if a is real and positive, then the result of the integral is
(Pi/2) a^n Hypergeometric2F1[(1-n)/2,-n/2,2,1/a^2]

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This is an interesting problem for sure. Concerning the integral values, the formulas look nice; they are Pi muliplied by a polynomial of alpha (of degree n). The coefficient of a^n is (1/2). The constant term in the polynomial is zero for odd values of n. In n is even, only even powers of a appear, if n is odd, only odd powers of n appear. I have not been able to establish recurrence relations for the coefficients.

I know this is not the full answer to your question, but may be this could help you finding some tracks. In any manner, thanks for the problem !

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