Snags when discovering the asymptotic behavior of an integral

Question

I have trouble in discovering the asymptotic behavior (i.e, the asymptotic expansion) of the following integral:

$$\newcommand\abs[1]{\left\lvert#1\right\rvert} \int_0^{\pi/2}\frac{dx}{1+(n\pi+x)\sin x}\tag1$$

The motivation is about the convergence of such improper integral:

$$I=\int_0^\infty\frac{x^\beta dx}{1+x^\alpha\abs{\sin x}}\qquad\alpha,\beta>0$$

The preceding integral is due to G.Hardy, which could be solved in such a way:

$$I_n=\int_{n\pi}^{(n+1)\pi}\frac{x^\beta dx}{1+x^\alpha\abs{\sin x}}=\int_0^\pi\frac{(n\pi+x)^\beta dx}{1+(n\pi+x)^\alpha\sin x}$$

Therefore $I=\sum_n I_n$ and we only need to estimate the asymptotic behavior of $I_n$.

In fact, it could be estimate roughly, which is sufficient to determine the convergence of $I$:

$$\frac{(n\pi)^\beta}{1+((n+1)\pi)^\alpha\sin x}\le\frac{(n\pi+x)^\beta}{1+(n\pi+x)^\alpha\sin x}\le\frac{((n+1)\pi)^\beta}{1+(n\pi)^\alpha\sin x}\qquad 0\le x\le\pi$$

Both the leftest side and the rightest side could be integrated over the closed interval $[0,\pi]$ elementarily and they are asymptotically equivalent (approximately $Cn^{-1}\log n$ where $C$ is a constant).

I want to see closer to the preceding integral. Can we obtain the asymptotic expansion of $I_n$? Out of simplicity, take $\alpha=1$ and $\beta=0$, and with a slight modification, we'll deal with the simpler version (1).

I have tried Laplace's method but failed after some efforts. It seems that Laplace method could estimate $\int_0^{\pi/n}$ well, but $\int_{\pi/n}^{\pi/2}$ is not a negligible tail.

Answer 1

We write the integrand as

$$ \begin{align} \frac{1}{1+n\pi\sin x + x\sin x} &= \frac{1}{1+n\pi\sin x} \cdot \frac{1}{1+\frac{x\sin x}{1+n\pi\sin x}} \\ &=\frac{1}{1+n\pi\sin x} \left[1 + O\left(\frac{x\sin x}{1+n\pi\sin x}\right)\right], \end{align} $$

where the $O(\cdots)$ holds uniformly for $n\geq 1$ and $x \in [0,\pi/2]$, so that

$$ \begin{align} &\int_0^{\pi/2} \frac{dx}{1+n\pi\sin x + x\sin x} \\ &\qquad= \int_0^{\pi/2} \frac{dx}{1+n\pi\sin x} + O\left(\int_0^{\pi/2} \frac{x\sin x}{(1+n\pi\sin x)^2}\,dx\right). \tag{1} \end{align} $$

We can estimate the integral in the $O(\cdots)$ by observing that

$$ \begin{align} \int_0^{\pi/2} \frac{x\sin x}{(1+n\pi\sin x)^2}\,dx &= \frac{1}{n^2 \pi^2} \int_0^{\pi/2} \frac{x\sin x}{\left(\frac{1}{n\pi}+\sin x\right)^2}\,dx \\ &\sim \frac{1}{n^2 \pi^2} \int_0^{\pi/2} \frac{x}{\sin x}\,dx \end{align} $$

by the dominated convergence theorem.

In light of this, $(1)$ becomes

$$ \int_0^{\pi/2} \frac{dx}{1+n\pi\sin x + x\sin x} = \int_0^{\pi/2} \frac{dx}{1+n\pi\sin x} + O\left(\frac{1}{n^2}\right). \tag{2} $$

Now courtesy of Mathematica we have

$$ \int_0^{\pi/2} \frac{dx}{1+n\pi\sin x} = \frac{\log\left(n\pi+\sqrt{n^2\pi^2-1}\right)}{\sqrt{n^2\pi^2-1}}, $$

which allows us to conclude from $(2)$ that

$$ \int_0^{\pi/2} \frac{dx}{1+n\pi\sin x + x\sin x} = \frac{\log\left(n\pi+\sqrt{n^2\pi^2-1}\right)}{\sqrt{n^2\pi^2-1}} + O\left(\frac{1}{n^2}\right). \tag{3} $$

Here's a plot showing the numerical value of the integral in blue and the asymptotic in purple for $1 \leq n \leq 10$.

As discussed in the comments, we can estimate the error term precisely by noting that the integrand of the error is

$$ \begin{align} \frac{1}{1+n\pi\sin x+x\sin x} - \frac{1}{1+n\pi\sin x} &= \frac{1+n\pi\sin x - 1 - n\pi\sin x - x\sin x}{(1+n\pi\sin x+x\sin x)(1+n\pi\sin x)} \\ &= \frac{1}{n^2\pi^2} \cdot \frac{-x\sin x}{\left(\sin x + \frac{1+x\sin x}{n\pi}\right)\left(\sin x + \frac{1}{n\pi}\right)}, \end{align} $$

so that, by the dominated convergence theorem, the error is asymptotic to

$$ - \frac{1}{n^2\pi^2} \int_0^{\pi/2} \frac{x}{\sin x}\,dx = - \frac{2G}{n^2\pi^2}, $$

where $G$ is Catalan's constant. Thus

$$ \begin{align} &\int_0^{\pi/2} \frac{dx}{1+n\pi\sin x + x\sin x} \\ &\qquad = \frac{\log\left(n\pi+\sqrt{n^2\pi^2-1}\right)}{\sqrt{n^2\pi^2-1}} - \frac{2G}{n^2\pi^2} + o\left(\frac{1}{n^2}\right). \tag{4} \end{align} $$

Here's a plot of the numerical integral in blue and this new asymptotic in purple.