Approximation of the sum of a series $S(t)=-frac{2}{pi t} cos(frac{pi t}{2}) sum_{m odd}^{infty}frac{m^2alpha_m}{t^2-m^2}$ as $tto +infty$

The function S(t) has the following infinite series form:

begin{align}
S(t) &=frac 2pi 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^2alpha_m}{t^2-m^2} \
end{align}

Where $m=1,3,5,…$ (odd numbers), $alpha_m$ is given by this messy expression:

$$alpha_m=-frac{1}{2m}left[(frac ab+1)J_{frac{m-1}{2}}(frac{mepsilon}{2})+(frac ab-1)J_{frac{m+1}{2}}(frac{mepsilon}{2})right]e^{mU_b} + frac{1}{2m}left[(frac ab-1)J_{frac{m-1}{2}}(frac{mepsilon}{2})+(frac ab+1)J_{frac{m+1}{2}}(frac{mepsilon}{2})right]e^{-mU_b}$$

where $a$ and $b$ are some positive constants $(a>b>0)$, $J$ is the Bessel function of first kind, $epsilon=displaystyle frac{a^2-b^2}{a^2+b^2}$, $U_b=displaystylefrac 12[tanh(2mu_b)-2mu_b]$ and $mu_b=tanh^{-1}(displaystylefrac ba)$.

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:

$$alpha_m sim frac{p}{m^{3/2}} qquad asquad m to +infty$$

where $p$ is a positive constant.

My question is:
How can I get the asymptotic approximation of $S(t)$ as $t to +infty$ ?

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My observation and attempts:

1. 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^2alpha_m}{t^2-m^2}\
&=-frac{2}{pi t^3} cos(frac{pi t}{2}) sum_{m odd}^{infty}frac{m^2alpha_m}{1-(frac{m}{t})^2}
end{align}

Since $displaystylesum_{m odd}^{infty} m^2 alpha_m $ diverges due to the asymptotic behavior of $alpha_m$, if we look at the sum $displaystylesum_{m odd}^{infty}frac{m^2 alpha_m}{1-frac{m^2}{t^2}}$, 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 $displaystylefrac{p}{m^{3/2}}$ in the sum. Thus,

$$S(t)=-frac{2}{pi t} cos(frac{pi t}{2}) sum_{m odd}^{infty}frac{m^2alpha_m}{t^2-m^2} sim -frac{2p}{pi t} cos(frac{pi t}{2}) sum_{m odd}^{infty}frac{m^{1/2}}{t^2-m^2} quad as quad t to +infty$$

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 $t to +infty$.

3. 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 $cos(displaystylefrac{pi t}{2}) = cos[displaystylefrac{pi}{2}(t-m)+displaystylefrac pi2 m]=-sin[displaystylefrac{pi t}{2}(1-displaystylefrac mt)]sin(displaystylefrac{pi m}{2})$, 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 $sin(displaystylefrac{pi m}{2})$.

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!

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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.