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?
 A: 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:
$\mathcal{M}_s^{-1}\left[\int_0^{\infty } \mathcal{M}_t\left[\frac{p^2 \cos \left(\sqrt{m^2+p^2} t\right)}{2 \left(m^2+p^2\right)^{5/2}}\right](s) \,
   dp\right](t)=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \frac{1}{2} p^2 \left(m^2+p^2\right)^{-\frac{5}{2}-\frac{s}{2}} \cos \left(\frac{\pi  s}{2}\right) \Gamma (s) \,
   dp\right](t)=\\\mathcal{M}_s^{-1}\left[\frac{m^{-2-s} \sqrt{\pi } \cos \left(\frac{\pi  s}{2}\right) \Gamma \left(1+\frac{s}{2}\right) \Gamma (s)}{8 \Gamma
   \left(\frac{5+s}{2}\right)}\right](t)=\\\frac{\pi  G_{1,3}^{2,0}\left(\frac{m t}{2},\frac{1}{2}|
\begin{array}{c}
 \frac{5}{2} \\
 0,1,\frac{1}{2} \\
\end{array}
\right)}{8 m^2}$
A: I prefer to add another answer since based on a different approach.
$$I=\int_0^{\infty } \frac{p^2 \sin ^2\left(\frac{t}{2} \sqrt{m^2+p^2} \right)}{\left(m^2+p^2\right)^{5/2}} \, dp=\frac12\int_0^{\infty } \frac{p^2}{ \left(m^2+p^2\right)^{5/2}}\, dp-\frac12\int_0^{\infty } \frac{p^2 \cos \left(t\sqrt{m^2+p^2} \right)}{ \left(m^2+p^2\right)^{5/2}} \, dp$$
Let $p=m q$
$$2m ^2\,I=\int_0^{\infty }\frac{q^2}{\left(q^2+1\right)^{5/2}}\,dq-\int_0^{\infty }\frac{q^2 \cos \left(m t\sqrt{q^2+1}
   \right)}{\left(q^2+1\right)^{5/2}}\,dq$$
$$\frac 13-2m^2\,I=\int_0^{\infty }\frac{q^2 \cos \left(mt \sqrt{q^2+1}
   \right)}{\left(q^2+1\right)^{5/2}}\,dq$$
$$mt \sqrt{q^2+1}=x \implies q=\frac{\sqrt{x^2-m^2 t^2}}{m t}\implies dq=\frac{x}{m t \sqrt{x^2-m^2 t^2}}\,dx$$
$$\int_0^{\infty }\frac{q^2 \cos \left(mt \sqrt{q^2+1}
   \right)}{\left(q^2+1\right)^{5/2}}\,dq=(mt)^2 \int_{mt}^\infty \sqrt{x^2-m^2 t^2}\,\,\frac{\cos (x) }{x^4}\,dx$$
$$\frac{1-6m^2\,I}{(mt)^2}=\int_{mt}^\infty \sqrt{x^2-m^2 t^2}\,\,\frac{\cos (x) }{x^4}\,dx$$
$$\sqrt{x^2-m^2 t^2}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n} (m t)^{2 n}\, x^{1-2 n}$$
$$\frac{1-6m^2\,I}{(mt)^2}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n} (m t)^{2 n}\int_{mt}^\infty  \frac {\cos (x)}{x^{2 n+3} }\,dx$$
$$2 \left(6  m^2\,I-1\right)=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}\Big[E_{2 n+3}(+i m t)+E_{2 n+3}(-i m t)\Big]$$
