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?

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 $p=m q$
$$2m ^2\,I=\int_0^{\infty }\frac{q^2}{\left(q^2+1\right)^{5/2}}\,dq-\int_0^{\infty }\frac{q^2 \cos \left(m t\sqrt{q^2+1}
   \right)}{\left(q^2+1\right)^{5/2}}\,dq$$
$$\frac 13-2m^2\,I=\int_0^{\infty }\frac{q^2 \cos \left(mt \sqrt{q^2+1}
   \right)}{\left(q^2+1\right)^{5/2}}\,dq$$
$$mt \sqrt{q^2+1}=x \implies q=\frac{\sqrt{x^2-m^2 t^2}}{m t}\implies dq=\frac{x}{m t \sqrt{x^2-m^2 t^2}}\,dx$$
$$\int_0^{\infty }\frac{q^2 \cos \left(mt \sqrt{q^2+1}
   \right)}{\left(q^2+1\right)^{5/2}}\,dq=(mt)^2 \int_{mt}^\infty \sqrt{x^2-m^2 t^2}\,\,\frac{\cos (x) }{x^4}\,dx$$
$$\frac{1-6m^2\,I}{(mt)^2}=\int_{mt}^\infty \sqrt{x^2-m^2 t^2}\,\,\frac{\cos (x) }{x^4}\,dx$$
$$\sqrt{x^2-m^2 t^2}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n} (m t)^{2 n}\, x^{1-2 n}$$
$$\frac{1-6m^2\,I}{(mt)^2}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n} (m t)^{2 n}\int_{mt}^\infty  \frac {\cos (x)}{x^{2 n+3} }\,dx$$
$$2 \left(6  m^2\,I-1\right)=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}\Big[E_{2 n+3}(+i m t)+E_{2 n+3}(-i m t)\Big]$$
