The function S(t) has the following infinite series form:
\begin{align} S(t) &=\frac 2\pi \int_0^{\pi/2} dx \sin(tx){\sum_{m\ odd}^{\infty} \alpha_m \cos[m(\frac \pi2-x)]}\\ &=-\frac{2}{\pi t} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{m^2\alpha_m}{t^2-m^2} \\ \end{align}
Where $m=1,3,5,...$ (odd numbers), $\alpha_m$ is given by this messy expression:
$$\alpha_m=-\frac{1}{2m}\left[(\frac ab+1)J_{\frac{m-1}{2}}(\frac{m\epsilon}{2})+(\frac ab-1)J_{\frac{m+1}{2}}(\frac{m\epsilon}{2})\right]e^{mU_b} + \frac{1}{2m}\left[(\frac ab-1)J_{\frac{m-1}{2}}(\frac{m\epsilon}{2})+(\frac ab+1)J_{\frac{m+1}{2}}(\frac{m\epsilon}{2})\right]e^{-mU_b}$$
where $a$ and $b$ are some positive constants $(a>b>0)$, $J$ is the Bessel function of first kind, $\epsilon=\displaystyle \frac{a^2-b^2}{a^2+b^2}$, $U_b=\displaystyle\frac 12[\tanh(2\mu_b)-2\mu_b]$ and $\mu_b=\tanh^{-1}(\displaystyle\frac ba)$.
I don't think this explicit form of $\alpha_m$ is going to help much in this problem, however, it has a very neat asymptotic behavior as:
$$\alpha_m \sim \frac{p}{m^{3/2}} \qquad as\quad m \to +\infty$$
where $p$ is a positive constant.
My question is:
How can I get the asymptotic approximation of $S(t)$ as $t \to +\infty$ ?
My observation and attempts:
If $t \to m'$, $m'$ is some large odd integer, we have: $$S(t) \to \frac{p}{2m'^{3/2}} \sin(\frac{\pi m'}{2})$$
\begin{align} S(t) &=-\frac{2}{\pi t} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{m^2\alpha_m}{t^2-m^2}\\ &=-\frac{2}{\pi t^3} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{m^2\alpha_m}{1-(\frac{m}{t})^2} \end{align}
Since $\displaystyle\sum_{m\ odd}^{\infty} m^2 \alpha_m $ diverges due to the asymptotic behavior of $\alpha_m$, if we look at the sum $\displaystyle\sum_{m\ odd}^{\infty}\frac{m^2 \alpha_m}{1-\frac{m^2}{t^2}}$, when $t \to +\infty$ the major contribution should come from $\alpha_m$ of large $m$. This, in some sense, justifies (not rigorously though) the move for replacing $\alpha_m$ by its asymptotic form $\displaystyle\frac{p}{m^{3/2}}$ in the sum. Thus,
$$S(t)=-\frac{2}{\pi t} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{m^2\alpha_m}{t^2-m^2} \sim -\frac{2p}{\pi t} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{m^{1/2}}{t^2-m^2} \quad as \quad t \to +\infty$$
Now, the goal is to find the asymptotics to $$I \equiv -\frac{2p}{\pi t} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{m^{1/2}}{t^2-m^2} $$ as $t \to +\infty$.
- My attempts to solve this problem were all to approximate the summation with an integral by writing summation in the form of Reimann sum as $t \to +\infty$.
For instance, my first try is as following:
\begin{align} I &\equiv -\frac{2p}{\pi t} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{m^{1/2}}{t^2-m^2}\\ &=-\frac{p}{\pi t^{\frac32}} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{(\frac mt)^{1/2}}{1-(\frac mt)^2} \frac 2t\\ &\sim -\frac{p}{\pi t^{\frac32}} \cos(\frac{\pi t}{2}) \int_0^{+\infty}\frac{x^{1/2}}{1-x^2}dx \quad as \quad t \to +\infty \end{align}
Since the integral $\int_0^{+\infty}\frac{x^{1/2}}{1-x^2}dx$ diverges due to the pole at $x=1$, but the principle value exists and $V.P. \int_0^{+\infty}\frac{x^{1/2}}{1-x^2}dx=-\frac \pi 2$, it follows that
$$I \sim \frac{p}{2 t^{\frac32}} \cos(\frac{\pi t}{2}) \quad as \quad t \to +\infty$$
However, approximating the well-behaved summation by a divergent integral seems a bit problematic, because in this way I did not include the effect of $\cos(\pi t/2)$ term which resolves the singularity problem and also it is inconsistent with my observation $1$.
Therefore my second attempt is to include $\cos(\pi t/2)$ in my summation. Using triogeometric property $\cos(\displaystyle\frac{\pi t}{2}) = \cos[\displaystyle\frac{\pi}{2}(t-m)+\displaystyle\frac \pi2 m]=-\sin[\displaystyle\frac{\pi t}{2}(1-\displaystyle\frac mt)]\sin(\displaystyle\frac{\pi m}{2})$, I thus deduce:
\begin{align} I &= -\frac{2p}{\pi t} \cos(\frac{\pi t}{2}) \sum_{m\ odd}^{\infty}\frac{m^{1/2}}{t^2-m^2}\\ &=-\frac{2p}{\pi t}\sum_{m\ odd}^{\infty}\frac{m^{1/2}}{t^2-m^2}\cos(\frac{\pi t}{2})\\ &=\frac{p}{\pi t^{\frac32}} \sum_{m\ odd}^{\infty}\frac{(\frac mt)^{1/2}}{1-(\frac mt)^2} \sin(\frac{\pi m}{2})sin[\frac{\pi t}{2}(1-\frac mt)]\frac 2t\\ &\sim \frac{p}{\pi t^{\frac32}} \int_0^{\infty} \frac{x^{1/2}}{1-x^2}\sin(\frac{\pi tx}{2})\sin[\frac{\pi t}{2}(1-x)]dx \quad as \quad t \to +\infty \end{align}
The above integral can be solved using methods suggested in the comments and answers of this question. Hence,
$$I \sim \frac{p}{4t^{\frac 32}} \sin(\frac{\pi t}{2}) + \frac{p}{4t^{\frac 32}} \cos(\frac{\pi t}{2}) \quad as \quad t \to +\infty$$
However, this result is still not consistent with my observation $1$ (I am short by a factor of $2$), also I think there is a problem in the step transforming summation into integral due to the term $\sin(\displaystyle\frac{\pi m}{2})$.
At this point, I don't know what should I do to proceed! Any tips, comments, and suggestions are very much welcomed and also I am willing to acknowledge anyone who provides any sorts of help to this problem in my work. Thank you all in advance!
Edit
I have figured out how to deal with this problem now, the key is indeed the Euler-Maclaurin formula. I am going to post the answer in case anyone is curious about it.