How might one tease out the asymptotic behaviour for large $x$ for the following function ($x$, $m$ both real and positive) $$s\left( x,m \right)=x\int\limits_{1}^{x}{\sin \left( \left( \frac{1}{x}+2 \right)m\pi \left( t+\frac{1}{t} \right) \right)dt}$$ For large $m$, I think it’s relatively straightforward. But for large $x$ I’ve had only limited success. In one approach, I expanded the sine and got coefficients on $x$ involving Bessel functions, but it requires an infinite series of them, the convergence of which I am unsure.
Another similar example is to consider estimating for large positive $x$, the integral $$\begin{align} & \operatorname{Im}\int\limits_{1}^{x}{{{e}^{\frac{i\left( 1+2x \right)\pi m}{t}}}dt}\underset{t\to 1/t}{\mathop{=}}\,\operatorname{Im}\int\limits_{1/x}^{1}{\frac{1}{{{t}^{2}}}{{e}^{i\left( 1+2x \right)\pi mt}}dt}\underset{s=tx}{\mathop{=}}\,\operatorname{Im}x\int\limits_{1}^{x}{\frac{1}{{{s}^{2}}}{{e}^{i\left( \frac{1}{x}+2 \right)\pi ms}}ds} \\ & =\operatorname{Im}\left\{ x\int\limits_{1}^{\infty }{\frac{1}{{{s}^{2}}}{{e}^{i\left( \frac{1}{x}+2 \right)\pi ms}}ds}-x\int\limits_{x}^{\infty }{\frac{1}{{{s}^{2}}}{{e}^{i\left( \frac{1}{x}+2 \right)\pi ms}}ds} \right\} \\ & \le \operatorname{Im}\left\{ x\int\limits_{1}^{\infty }{\frac{1}{{{s}^{2}}}{{e}^{i2\pi ms}}ds}+x\int\limits_{x}^{\infty }{\frac{1}{{{s}^{2}}}ds} \right\}=-2\pi m\operatorname{Ci}\left( 2\pi m \right)x+O\left( 1 \right) \\ \end{align}$$ where $\operatorname{Ci}$ is the cosine integral. This approximation seems quite good, but trying to apply this sort of reasoning to the first example runs into problems.
The following might possibly be a solution. For now consider the slightly simpler integral, where we assume $a$ is independent of $x$ $$j\left( x \right)=\int\limits_{1}^{x}{\sin \left( a\left( t+\frac{1}{t} \right) \right)dt}$$
It’s not difficult to then show, by first letting $t={{e}^{u}}$ and then $z=\cosh \left( u \right)$ $$j\left( x \right)=\frac{1}{2a}\left\{ \cos \left( 2a \right)-\cos \left( a\left( x+\frac{1}{x} \right) \right) \right\}+\int\limits_{1}^{\frac{1}{2}\left( x+\frac{1}{x} \right)}{\frac{z\sin \left( 2az \right)}{\sqrt{{{z}^{2}}-1}}dz}$$
Consider now $$\int\limits_{C}^{{}}{\frac{z\sin \left( 2az \right)}{\sqrt{{{z}^{2}}-1}}dz}$$
where the contour $C$ involves a quarter of a large circle of radius $\frac{1}{2}\left( x+\frac{1}{x} \right)$, starting on the positive real-axis, ending on the positive-imaginary axis, joined by two straight line segments, one along the imaginary axis, and one along the real-axis which involves a small (negligible) detour around the branch point $z=+1$.
We choose local polar coordinates for the cut such that
$$\sqrt{z+1}=r_{-1}^{1/2}{{e}^{i{{\theta }_{-1}}/2}},\,\,\sqrt{z-1}=r_{+1}^{1/2}{{e}^{i{{\theta }_{+1}}/2}},\ \ 0\le {{\theta }_{+/-1}}<2\pi $$
Taking care with the parameterisation we have
$$\begin{aligned} \int\limits_{1}^{\frac{1}{2}\left( x+\frac{1}{x} \right)}{\frac{t{{e}^{i2at}}}{\sqrt{{{t}^{2}}-1}}dt}&=i\int\limits_{0}^{1}{\frac{t{{e}^{i2at}}}{\sqrt{1-{{t}^{2}}}}dt}-\int\limits_{0}^{\pi /2}{\frac{\frac{1}{2}\left( x+\frac{1}{x} \right){{e}^{i\theta }}{{e}^{i2a\frac{1}{2}\left( x+\frac{1}{x} \right){{e}^{i\theta }}}}}{\sqrt{\frac{1}{4}{{\left( x+\frac{1}{x} \right)}^{2}}{{e}^{2i\theta }}-1}}\frac{1}{2}\left( x+\frac{1}{x} \right)i{{e}^{i\theta }}d\theta }\\&+i\int\limits_{0}^{\frac{1}{2}\left( x+\frac{1}{x} \right)}{\frac{t{{e}^{-2at}}}{\sqrt{{{t}^{2}}+1}}dt}\end{aligned}$$
Now first consider
$$\begin{aligned} &\int\limits_{0}^{\pi /2}{\frac{\frac{1}{2}\left( x+\frac{1}{x} \right){{e}^{i\theta }}{{e}^{ia\left( x+\frac{1}{x} \right){{e}^{i\theta }}}}}{\sqrt{\frac{1}{4}{{\left( x+\frac{1}{x} \right)}^{2}}{{e}^{2i\theta }}-1}}\frac{1}{2}\left( x+\frac{1}{x} \right)i{{e}^{i\theta }}d\theta }\\&=\sum\limits_{n=0}^{\infty }{\frac{{{\left( ia \right)}^{n}}}{n!{{x}^{n}}}}\int\limits_{0}^{\pi /2}{{{e}^{iax{{e}^{i\theta }}+i2\theta +in\theta }}\left\{ \frac{1}{2}i{{e}^{-i\theta }}x+\frac{i{{e}^{-i3\theta }}}{2x}\left( 2+{{e}^{2i\theta }} \right)+O\left( {{x}^{-3}} \right) \right\}d\theta }\end{aligned}$$
Where we have expanded main arguments of the integrand (including the exponential involving $1/x$) in powers of x. Collecting terms
$$\begin{aligned} &\int\limits_{0}^{\pi /2}{\frac{\frac{1}{2}\left( x+\frac{1}{x} \right){{e}^{i\theta }}{{e}^{i2a\frac{1}{2}\left( x+\frac{1}{x} \right){{e}^{i\theta }}}}}{\sqrt{\frac{1}{4}{{\left( x+\frac{1}{x} \right)}^{2}}{{e}^{2i\theta }}-1}}\frac{1}{2}\left( x+\frac{1}{x} \right)i{{e}^{i\theta }}d\theta }\\&=\int\limits_{0}^{\pi /2}{{{e}^{iax{{e}^{i\theta }}+i2\theta }}\left\{ \frac{1}{2}i{{e}^{-i\theta }}\left( x+\frac{1}{x} \right)-\frac{1}{2}a-i\frac{{{a}^{2}}{{e}^{i\theta }}}{4x}+\frac{i{{e}^{-i3\theta }}}{x} \right\}d\theta }+O\left( {{x}^{-2}} \right)\end{aligned}$$
The integrals for the first three terms can be handled in the following manner
$$\int\limits_{0}^{\pi /2}{{{e}^{iax{{e}^{i\theta }}+i2\theta }}\left\{ \frac{1}{2}i{{e}^{-i\theta }}\left( x+\frac{1}{x} \right)-\frac{1}{2}a-i\frac{{{a}^{2}}{{e}^{i\theta }}}{4x} \right\}d\theta }=\int\limits_{1}^{i}{{{e}^{iaxt}}\left\{ \frac{1}{2}\left( x+\frac{1}{x} \right)-\frac{{{a}^{2}}{{t}^{2}}}{4x}+\frac{i}{2}ta \right\}dt}$$
The justification for this is that we go back into the complex plane, and create a quarter circle loop, i.e.
$$\int\limits_{C}^{{}}{{}}=\int\limits_{0}^{1}{{}}+\int\limits_{0}^{\pi /2}{{}}+\int\limits_{i}^{0}{{}}=0\Rightarrow +\int\limits_{0}^{\pi /2}{{}}=\int\limits_{1}^{0}{{}}+\int\limits_{0}^{i}{{}}\equiv \int\limits_{1}^{i}{{}}$$
These are relatively easy via integration by parts, and so
$$\begin{align} & \int\limits_{1}^{i}{{{e}^{iaxt}}\left\{ \frac{1}{2}\left( x+\frac{1}{x} \right)-\frac{{{a}^{2}}{{t}^{2}}}{4x}+\frac{i}{2}ta \right\}dt} \\ & =\frac{i}{2a{{x}^{4}}}\left( {{e}^{iax}}-{{e}^{-ax}} \right)+\frac{1}{2{{x}^{3}}}\left( {{e}^{iax}}-i{{e}^{-ax}} \right)-\frac{ia}{4{{x}^{2}}}\left( {{e}^{iax}}+{{e}^{-ax}} \right)+\frac{1}{2x}\left( i{{e}^{-ax}}-{{e}^{iax}} \right)+\frac{i}{2a}\left( {{e}^{iax}}-{{e}^{-ax}} \right) \\ \end{align}$$
For the remaining integral consider then
$$\int\limits_{0}^{\pi /2}{{{e}^{iax{{e}^{i\theta }}+i2\theta }}\left\{ \frac{i{{e}^{-i3\theta }}}{x} \right\}d\theta }=\frac{i}{x}\int\limits_{0}^{\pi /2}{{{e}^{iax{{e}^{i\theta }}-i\theta }}d\theta }=\frac{i}{x}\int\limits_{0}^{\pi /2}{{{e}^{iax\cos \left( \theta \right)-i\theta }}{{e}^{-ax\sin \left( \theta \right)}}d\theta }$$
For large $x$, the bulk contribution comes from near $\theta \simeq 0$ and so we have via standard asymptotic methods
$$\int\limits_{0}^{\pi /2}{{{e}^{iax{{e}^{i\theta }}+i2\theta }}\left\{ \frac{i{{e}^{-i3\theta }}}{x} \right\}d\theta }=\frac{i}{x}\int\limits_{0}^{\pi /2}{{{e}^{iax{{e}^{i\theta }}-i\theta }}d\theta }\sim \frac{i}{x}\int\limits_{0}^{\infty }{{{e}^{iax}}{{e}^{-ax\theta }}d\theta }=\frac{i{{e}^{iax}}}{a{{x}^{2}}}=O\left( {{x}^{-2}} \right)$$
We find then
$$\int\limits_{0}^{\pi /2}{\frac{\frac{1}{2}\left( x+\frac{1}{x} \right){{e}^{i\theta }}{{e}^{i2a\frac{1}{2}\left( x+\frac{1}{x} \right){{e}^{i\theta }}}}}{\sqrt{\frac{1}{4}{{\left( x+\frac{1}{x} \right)}^{2}}{{e}^{2i\theta }}-1}}\frac{1}{2}\left( x+\frac{1}{x} \right)i{{e}^{i\theta }}d\theta }=\frac{1}{2x}\left( i{{e}^{-ax}}-{{e}^{iax}} \right)+\frac{i}{2a}\left( {{e}^{iax}}-{{e}^{-ax}} \right)+O\left( {{x}^{-2}} \right)$$
Next
$$\int\limits_{0}^{1}{\frac{t{{e}^{i2at}}}{\sqrt{1-{{t}^{2}}}}dt}=\frac{1}{2i}\frac{d}{da}\int\limits_{0}^{1}{\frac{{{e}^{i2at}}}{\sqrt{1-{{t}^{2}}}}dt}=\frac{1}{2i}\frac{d}{da}\left\{ \int\limits_{0}^{1}{\frac{\cos \left( 2at \right)}{\sqrt{1-{{t}^{2}}}}dt}+i\int\limits_{0}^{1}{\frac{\sin \left( 2at \right)}{\sqrt{1-{{t}^{2}}}}dt} \right\}$$
Note, by definition
$$\int\limits_{0}^{1}{\frac{\cos \left( 2at \right)}{\sqrt{1-{{t}^{2}}}}dt}=\frac{1}{2}\pi {{J}_{0}}\left( 2a \right),\ \ \int\limits_{0}^{1}{\frac{\sin \left( 2at \right)}{\sqrt{1-{{t}^{2}}}}dt}=\frac{1}{2}\pi {{\mathbf{H}}_{0}}\left( 2a \right)$$
where we have the Bessel and Struve functions of the first kind. Thus
$$\int\limits_{0}^{1}{\frac{t{{e}^{i2at}}}{\sqrt{1-{{t}^{2}}}}dt}=\frac{\pi }{4}\frac{d}{da}\left\{ {{\mathbf{H}}_{0}}\left( 2a \right)-i{{J}_{0}}\left( 2a \right) \right\}$$
And noting
$$\frac{d}{dz}{{\mathbf{H}}_{0}}\left( z \right)={{\mathbf{H}}_{-1}}\left( z \right),\ \ \frac{d}{dz}{{J}_{0}}\left( z \right)=-{{J}_{1}}\left( z \right)$$
We conclude $$i\int\limits_{0}^{1}{\frac{t{{e}^{i2at}}}{\sqrt{1-{{t}^{2}}}}dt}=\frac{\pi }{2}\left\{ -{{J}_{1}}\left( 2a \right)+i{{\mathbf{H}}_{-1}}\left( 2a \right) \right\}$$ Lastly $$\begin{align} & \int\limits_{0}^{\frac{1}{2}\left( x+\frac{1}{x} \right)}{\frac{t{{e}^{-2at}}}{\sqrt{{{t}^{2}}+1}}dt}=\int\limits_{0}^{\infty }{\frac{t{{e}^{-2at}}}{\sqrt{{{t}^{2}}+1}}dt}-\int\limits_{\frac{1}{2}\left( x+\frac{1}{x} \right)}^{\infty }{\frac{t{{e}^{-2at}}}{\sqrt{{{t}^{2}}+1}}dt} \\ & \le \int\limits_{0}^{\infty }{\frac{t{{e}^{-2at}}}{\sqrt{{{t}^{2}}+1}}dt}+\int\limits_{\frac{1}{2}\left( x+\frac{1}{x} \right)}^{\infty }{{{e}^{-2at}}dt}=\int\limits_{0}^{\infty }{\frac{t{{e}^{-2at}}}{\sqrt{{{t}^{2}}+1}}dt}+\frac{1}{2a}{{e}^{-a\left( x+\frac{1}{x} \right)}} \\ \end{align}$$ And we have also $$\int\limits_{0}^{\infty }{\frac{t{{e}^{-2at}}}{\sqrt{1+{{t}^{2}}}}dt}=-\frac{1}{2}\frac{d}{da}\int\limits_{0}^{\infty }{\frac{{{e}^{-2at}}}{\sqrt{1+{{t}^{2}}}}dt}$$ Again by definition $$\int\limits_{0}^{\infty }{\frac{{{e}^{-2at}}}{\sqrt{1+{{t}^{2}}}}dt}=\frac{\pi }{2}{{\mathbf{K}}_{0}}\left( 2a \right)=\frac{\pi }{2}\left\{ {{\mathbf{H}}_{0}}\left( 2a \right)-{{Y}_{0}}\left( 2a \right) \right\}$$ where Y is the Bessel function of the second kind. Using well known derivatives we find $$\int\limits_{0}^{\frac{1}{2}\left( x+\frac{1}{x} \right)}{\frac{t{{e}^{-2at}}}{\sqrt{{{t}^{2}}+1}}dt}=-\frac{\pi }{2}\left\{ {{\mathbf{H}}_{-1}}\left( 2a \right)+{{Y}_{1}}\left( 2a \right) \right\}+O\left( {{e}^{-ax}} \right)$$ Putting it all together $$\begin{aligned} &\int\limits_{1}^{\frac{1}{2}\left( x+\frac{1}{x} \right)}{\frac{t{{e}^{i2at}}}{\sqrt{{{t}^{2}}-1}}dt}\\&=\frac{\pi }{2}\left\{ -{{J}_{1}}\left( 2a \right)+i{{\mathbf{H}}_{-1}}\left( 2a \right) \right\}+\frac{{{e}^{iax}}}{2x}-\frac{i{{e}^{iax}}}{2a}-i\frac{\pi }{2}\left\{ {{\mathbf{H}}_{-1}}\left( 2a \right)+{{Y}_{1}}\left( 2a \right) \right\}+O\left( {{x}^{-2}} \right)\end{aligned}$$ Hence $$\int\limits_{1}^{\frac{1}{2}\left( x+\frac{1}{x} \right)}{\frac{t\sin \left( 2at \right)}{\sqrt{{{t}^{2}}-1}}dt}=-\frac{\pi }{2}{{Y}_{1}}\left( 2a \right)-\frac{\cos \left( ax \right)}{2a}+\frac{\sin \left( ax \right)}{2x}+O\left( {{x}^{-2}} \right)$$ And so $$\begin{aligned} j\left( x \right)&=\int\limits_{1}^{x}{\sin \left( a\left( t+\frac{1}{t} \right) \right)dt}\\&=\frac{1}{2a}\left\{ \cos \left( 2a \right)-\cos \left( a\left( x+\frac{1}{x} \right) \right) \right\}-\frac{\pi }{2}{{Y}_{1}}\left( 2a \right)-\frac{\cos \left( ax \right)}{2a}+\frac{\sin \left( ax \right)}{2x}+O\left( {{x}^{-2}} \right)\end{aligned}$$
$a=4$, blue = numerical, orange = asymptotic, green = difference of the two
Now we can use the same process for the main integral in question, its just we need to re-calculate our expansions in terms of $x$ given the coefficient $a$ now depends on $x$. For example we have in this instance
$$\int\limits_{0}^{\pi /2}{\frac{\frac{1}{2}\left( x+\frac{1}{x} \right){{e}^{i\theta }}{{e}^{i\left( \frac{1}{x}+2 \right)\pi m\left( x+\frac{1}{x} \right){{e}^{i\theta }}}}}{\sqrt{\frac{1}{4}{{\left( x+\frac{1}{x} \right)}^{2}}{{e}^{2i\theta }}-1}}\frac{1}{2}\left( x+\frac{1}{x} \right)i{{e}^{i\theta }}d\theta }=\int\limits_{0}^{\pi /2}{{{e}^{i\left( 1+2x \right)\pi m{{e}^{i\theta }}+i2\theta }}\left\{ \frac{i{{e}^{-i\theta }}}{2}\left( x+\frac{1}{x} \right)-\frac{\pi m}{2x}+\frac{i{{e}^{-i3\theta }}}{x}-\pi m-\frac{i{{\pi }^{2}}{{m}^{2}}{{e}^{i\theta }}}{x}+O\left( {{x}^{-2}} \right) \right\}d\theta }$$
And so $$\int\limits_{1}^{\frac{1}{2}\left( x+\frac{1}{x} \right)}{\frac{t\sin \left( \left( \frac{1}{x}+2 \right)2\pi mt \right)}{\sqrt{{{t}^{2}}-1}}dt}=-\frac{\pi }{2}{{Y}_{1}}\left( \left( \frac{1}{x}+2 \right)2\pi m \right)-\frac{\cos \left( \left( 1+2x \right)\pi m \right)}{4\pi m}\left( 1-\frac{1}{2x} \right)+\frac{\sin \left( \left( 1+2x \right)\pi m \right)}{2x}+O\left( {{x}^{-2}} \right)$$
We find then
$$\begin{align} & s\left( x,m \right)=x\int\limits_{1}^{x}{\sin \left( \left( \frac{1}{x}+2 \right)\pi m\left( t+\frac{1}{t} \right) \right)dt}= \\ & \frac{x}{\left( \frac{1}{x}+2 \right)2\pi m}\left\{ \cos \left( \left( \frac{1}{x}+2 \right)2\pi m \right)-\cos \left( \left( \frac{1}{x}+2 \right)\left( x+\frac{1}{x} \right)\pi m \right) \right\} \\ & -\frac{\pi }{2}x{{Y}_{1}}\left( \left( \frac{1}{x}+2 \right)2\pi m \right)-\frac{\cos \left( \left( 1+2x \right)\pi m \right)}{4\pi m}\left( x-\frac{1}{2} \right)+\frac{\sin \left( \left( 1+2x \right)\pi m \right)}{2}+O\left( {{x}^{-1}} \right) \\ \end{align}$$
For example consider the comparison for the case of $m=2$
$m=2$, blue=numerical (Levin method), orange = asymptotic (orange), green = difference between them
We can of course expand the Bessel function, along with other terms involving $x$, and doing so we’re left with an approximation, for large $x$, of the form
$$s\left( x,m \right)=\alpha \left( x,m \right)x+\beta \left( x,m \right)+O\left( {{x}^{-1}} \right)$$
where the coefficients are reasonably involved and so have been omitted here. But it should be noted that the coefficients have a dependence on $x$ within circular functions and so we can therefore estimate mean growth and bounds. For example
$$\left\langle s\left( x,2 \right) \right\rangle =\left\{ \frac{1}{8\pi }-\frac{\pi }{2}{{Y}_{1}}\left( 8\pi \right) \right\}x+2{{\pi }^{2}}{{Y}_{2}}\left( 8\pi \right)-\frac{\pi }{4}{{Y}_{1}}\left( 8\pi \right)-\frac{1}{16\pi }+O\left( {{x}^{-1}} \right)$$
$m=2$, blue = numerical, orange = asymptotic mean