What is the correct value of $\int_0^\infty dx\ J_0\left( 2 a \sinh\left( \frac{x}{2} \right) \right) \sin(b x)$? Integral (6.679.4) in Gradshteyn and Ryzhik claims that
$$
\int_0^\infty dx\ J_0\left( 2 a \sinh\left( \frac{x}{2} \right) \right) \sin(b x) \ = \ \frac{2}{\pi} \sinh(\pi b) \left[ K_{ib}(a) \right]^2
$$
I think that this is not correct. Here is a screenshot of some numerics in Mathematica:

For random values $a=2.34$ and $b=3$, it seems that the LHS of the above evaluates to $0.408$ while the RHS evaluates to $0.653$, and so they disagree.
I think there are three options:
1. I am making a mistake somehow in the above.
2. Mathematica is defining Bessel functions $J$ or $K$ differently than G&R, which I think is not likely.
2. This is an error in G&R.
In the last case, I have seen that Equation (58) on Page 115 of Erdelyi: Volume 1 Tables of Integral Transforms is the source for G&R (6.679.4). This integral simply stated there without a source, so I have no clue how this integral is derived.
What is $\int_0^\infty dx\ J_0\left( 2 a \sinh\left( \frac{x}{2} \right) \right) \sin(b x)$?
 A: Maple agrees with G&R on the formula, however it has considerable difficulty with the numerical evaluation of the integral as this oscillates very rapidly. But the integral from $0$ to $20$ is given as $.6536634068$, agreeing quite nicely with the theoretical result. 
A: Interesting problem, for sure.
Using another CAS, I played with the precision. For Mathematica, your  first command would be 


f[a_,b_,n_]:=NIntegrate[BesselJ[0, 2 a Sinh[x/2]] Sin[b x], {x, 0, Infinity}, 
     WorkingPrecision -> n]


Below are my results for the integration up to infinity for your particular values.
$$\left(
\begin{array}{cc}
  n & \text{result} \\
 10 & 0.4084616144 \\
 20 & 0.4084616143 \\
 30 & 0.6848603276 \\
 40 & 0.6848603276 \\
 50 & 0.6554693076 \\
 60 & 0.6554693076 \\
 70 & 0.6554693076 \\
 80 & 0.6554693076 \\
 90 &  0.6541206739\\
 100 & 0.6541738396 \\
 200 & 0.6533938298 \\
 300 & 0.6534598266 \\
 400 & 0.6536447955 \\
 500 & 0.6534985760
\end{array}
\right)$$ It seems to be very slow convergence (notice the swings of these results even for large values of $n$).
The systematic complain of the CAS was related to the highly oscillatory integrand as Robert Israel commented.
