Show that $$ \int_0^{\infty} kF(k)\sin(ka)\,dk = \frac{\pi}{2}aG(a) $$ where $$ F(x) = \frac{1}{2}+\frac{1-x^2}{4x}\ln\vert\frac{1+x}{1-x}\vert $$ and $$ G(x) = \frac{\sin x-x\cos x}{x^4} $$
EDIT: The source can be found here. One should notice that the function $F(x)$ is not continuous at $x=1$.
EDIT2: The integral below may be of some help. $$ \ln\vert\frac{a+x}{a-x}\vert = 2\int_0^{\infty}\frac{\sin at\sin xt}{t}\,dt $$
EDIT3: Here is a mathematica code of my question:
F[x_] := (1/2 + (1 - x^2)/(4 x)*Log[Abs[(1 + x)/(1 - x)]])xSin[a*x];
Integrate[F[x], {x, 0, Infinity}, Assumptions -> a > 0]
The result is:
([Pi] (-a Cos[a] + Sin[a]))/(2 a^3)
EDIT4: By virtue of Bessel function $J_{\frac{1}{2}}(z)=\sqrt{\frac{2}{\pi z}}\sin(z)$, the integral identity turns to be $$ \int_0^{\infty} k^{\frac{3}{2}}F(k)J_{\frac{1}{2}}(ka)\,dk = \sqrt{\frac{\pi}{2}a}G(a) $$ In this occasion, some refs are useful:
(1) Lin Q.G.: Infinite integrals involving Bessel functions by an improved approach of contour integration and the residue theorem.
(2) Lucas S.K.: Evaluating infinite integrals involving Bessel functions of arbitrary order.