Conjecture:$\int_{0}^{\infty} \frac{\sin^3(x)\sin(3\pi x)}{x^4} J_0(x)^2\text{d}x \overset{?}{=} \frac{\pi}{2}$ I found
$$\int_{0}^{\infty} \frac{\sin^3(x)\sin(3\pi x)}{x^4} J_0(x)^2\text{d}x=1.570796...
\overset{?}{=} \frac{\pi}{2}$$
Where $J_0(x)$ is the 0th Bessel functions of the first kind.
I try to use the following identity,
$$\int_{-\infty}^{\infty} J_0(x)e^{-i\omega x}\text{d}x
=\frac{1}{\sqrt{1-\omega^2} }$$
And
$$\int_{-\infty}^{\infty} f(x)g(x)e^{i\omega x}\text{d}x
=\frac{1}{2\pi} \int_{-\infty}^{\infty} F(x)G(\omega-x)\text{d}x$$
Where $F(x)=\int_{-\infty}^{\infty} f(t)e^{-itx}\text{d}t,
G(x)=\int_{-\infty}^{\infty} g(t)e^{-itx}\text{d}t$.
But the integral $\int_{-\infty}^{\infty} J_0(x)^2e^{i\omega x}\text{d}x$ involves $K(k)$(Complete elliptic integral of the first kind).So it's not a good way to prove it.
Now I don't have any idea to prove it. Are there any hints? Or what techniques should I use?
 A: This integral is sort of like a Borwein integral.
The function $$f(z) = \sin^{3}(z) e^{iaz} J_{0}(z)^{2}$$ is entire.
And as $|z| \to \infty$,the Bessel function of the first kind of order zero has the asymptotic form$$J_{0}(z) \sim \sqrt{\frac{2}{\pi z}} \cos \left(z-\frac{\pi}{4} \right) , \quad |\arg(z)| < \pi, $$ which means that $|f(z)|$ is bounded in the upper half of the complex plane if $a \ge 5$.
(If $ a<5$, $|f(z)|$ will grow exponentially as $\Im(z) \to +\infty$. And  if $  a > -5$, $|f(z)|$ will grow exponentially as $\Im(z) \to - \infty$.)
Therefore, if $a \ge 5$, we can integrate the function $$g(z) = \frac{f(z)}{z^{4}}$$ around an  infinitely large semicircular contour in the upper half-plane that is indented at the origin, apply the estimation lemma,  and conclude that $$ \operatorname{PV}\int_{-\infty}^{\infty} g(x) \, \mathrm dx- i \pi \operatorname{Res} \left[g(z), 0 \right] =0, $$ where $$\operatorname{Res} \left[g(z), 0 \right] = \lim_{z \to 0} \frac{\sin^{3}(z)}{z^{3}} \, e^{iaz} J_{0}(z)^{2}= 1(1)(1) =1. $$
Equating the imaginary parts on both sides of the equation, we get $$\int_{-\infty}^{\infty} \frac{\sin^{3}(x) \sin(ax) J_{0}(x)^{2}}{x^{4}} \, \mathrm dx = \pi \, , \quad a \ge 5.$$
Since the integrand is even, it follows that $$\int_{0}^{\infty} \frac{\sin^{3}(x) \sin(ax) J_{0}(x)^{2}}{x^{4}} \, \mathrm dx = \frac{\pi}{2} \, , \quad a \ge 5. $$
