Does the following definite integral exist? I encounter a problem  in which I would need to deal with the folloing definite integral
$$I(t)=\int_{0}^{\infty} \mathrm dp \frac{p^2}{\omega^5}  \sin^2\left(\frac{\omega t}{2}\right)$$
in which $$\omega=\sqrt{m^2+p^2}$$
where m is just some positive constant, t is also a parameter. But I would like to obtain the value of $I(t)$ when t is small. My approach to this problem is the following
$$I(t)=I(0)+I(0)'t+\frac{I(0)''}{2}t^2...$$
$$\frac{dI}{dt}|_{t=0}=0$$
However
$$I(0)''\propto \int dp \frac{p^2}{\omega^3}$$
which diverge like $\ln(p)$
Can anyone help me to find the expansion of I(t) in terms of t?

With Mathematica I have:
$$\int_0^{\infty } \frac{p^2 \sin ^2\left(\frac{1}{2} \sqrt{m^2+p^2} t\right)}{\left(m^2+p^2\right)^{5/2}} \, dp=\int_0^{\infty } \left(\frac{p^2}{2 \left(m^2+p^2\right)^{5/2}}-\frac{p^2 \cos \left(\sqrt{m^2+p^2} t\right)}{2 \left(m^2+p^2\right)^{5/2}}\right) \, dp=\\\frac{1}{6 m^2}-\frac{\pi  G_{1,3}^{2,0}\left(\frac{m
   t}{2},\frac{1}{2}|
\begin{array}{c}
 \frac{5}{2} \\
 0,1,\frac{1}{2} \\
\end{array}
\right)}{8 m^2}$$
for: $t>0$
MMA code:
Integrate[(p^2 Sin[1/2 Sqrt[m^2 + p^2] t]^2)/(m^2 + p^2)^( 5/2), {p, 0, \[Infinity]}] ==  1/(6 m^2) - (\[Pi] MeijerG[{{}, {5/2}}, {{0, 1}, {1/2}}, (m t)/2, 1/ 2])/(8 m^2)
Using MellinTransfrom:
$\mathcal{M}_s^{-1}\left[\int_0^{\infty } \mathcal{M}_t\left[\frac{p^2 \cos \left(\sqrt{m^2+p^2} t\right)}{2 \left(m^2+p^2\right)^{5/2}}\right](s) \,
   dp\right](t)=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \frac{1}{2} p^2 \left(m^2+p^2\right)^{-\frac{5}{2}-\frac{s}{2}} \cos \left(\frac{\pi  s}{2}\right) \Gamma (s) \,
   dp\right](t)=\\\mathcal{M}_s^{-1}\left[\frac{m^{-2-s} \sqrt{\pi } \cos \left(\frac{\pi  s}{2}\right) \Gamma \left(1+\frac{s}{2}\right) \Gamma (s)}{8 \Gamma
   \left(\frac{5+s}{2}\right)}\right](t)=\\\frac{\pi  G_{1,3}^{2,0}\left(\frac{m t}{2},\frac{1}{2}|
\begin{array}{c}
 \frac{5}{2} \\
 0,1,\frac{1}{2} \\
\end{array}
\right)}{8 m^2}$
