Integrating $\int_0^\infty \frac{\sin (ax)}{x^3}dx$ I am trying to integrate this integral,
$$I_3 = \frac{ -4 }{ \pi} \int_0 ^{\infty} \frac{d\lambda}{\lambda^4} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda) \left(-p_3 \lambda\cos\left(p_3 \lambda\right)\right)\cdot\sin\left(p_4 \lambda\right) \\
    = \frac{4p_3 }{ \pi} \int_0 ^{\infty} \frac{d\lambda}{\lambda^3} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda) \cos\left(p_3 \lambda\right)\sin\left(p_4 \lambda\right)$$
I changed this trigonometric product to a sum of  $\sin\left(\left(dp_4+cp_3+bp_2+ap_1\right)\lambda\right)$ where $a,b,c,d= \pm1$ But then integral takes the form $\int_0 ^{\infty} \frac{d\lambda}{\lambda^4} \sin\left(k\lambda\right)$ Which seems divergent to me. Is there any other way to integrate this?
Background: I am trying to calculate $$ D_2(p_3,p_4) =\frac{-4}{\pi} \int \frac{d\lambda}{\lambda^4} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda)
 \left(\sin\left(p_3 \lambda\right)-p_3 \lambda\cos\left((p_3 \lambda\right)\right)\cdot\left(\sin\left(p_4 \lambda\right)-p_4 \lambda\cos\left((p_4 \lambda\right)\right)$$ which has a finite value according to Appendix C of this paper and $I_3$ is one of the four terms after expanding it. If you can suggest any other simpler way to calculate $D_2$, that would also be great.
Thanks in advance :)
 A: $$\int\frac{\sin (ax)}{x^4}dx=a^3 \int\frac{\sin (t)}{t^4}dt$$ Now, you need a few integrations by parts to get
$$\int\frac{\sin (t)}{t^4}dt=-\frac{\cos (t)}{6 t^2}+\frac{\left(t^2-2\right) \sin (t)}{6 t^3}-\frac{\text{Ci}(t)}{6}$$
$$I(\epsilon)=\int_\epsilon^\infty\frac{\sin (t)}{t^4}dt=\frac{\cos (\epsilon)}{6 \epsilon^2}-\frac{\left(\epsilon^2-2\right) \sin (\epsilon)}{6 \epsilon^3}+\frac{\text{Ci}(\epsilon)}{6}$$
$$I(\epsilon)=\frac{1}{2 \epsilon ^2}+\frac{1}{36} (6 \log (\epsilon )+6 \gamma -11)-\frac{\epsilon
   ^2}{240}+\frac{\epsilon ^4}{20160}+O\left(\epsilon ^6\right)$$
A: $$I(a)=\int_0^\infty\frac{\sin(ax)}{x^4}dx$$
we can analyse its converge by the following: $u=ax\Rightarrow x=u/a,\,dx=du/a$
$$I(a)=\int_0^\infty\frac{\sin(u)}{(u/a)^4}\frac{du}a=a^3\int_0^\infty\frac{\sin(u)}{u^4}du$$
now we need to look at what happens as the integrand tends to zero, to do this I will split it into two parts:
$$J=\int_0^\epsilon\frac{\sin(u)}{u^4}du+\int_\epsilon^\infty\frac{\sin(u)}{u^4}du$$
the problem is this first integral, for a small enough $\epsilon$ it is fair for us to say $\sin u\approx u$ so:
$$\int_0^\epsilon\frac{\sin u}{u^4}du=\int_0^\epsilon\frac1{u^3}du\to\infty$$
