Fourier cosine transform from Erdélyi's Tables of Integral Transforms I’d like confirmation that
$$ \frac{\cos⁡(b \sqrt{a^2+x^2})}{(a^2+x^2)^{3/2}} $$
has the Fourier cosine transform
$$ \frac{\pi}{2 a} \, \exp(-ay) \qquad \text{if $y>a$,}
$$
as found in Tables of Integral Transforms by Arthur Erdelyi et al. equation (35) in Sect. 1.7.
I am puzzled that it is independent of $b$, making me wonder whether the inequality should be $y>b$. Since the numerator is a kind of frequency swing when $x$ is time, I would expect the spectrum to reflect the width $b$ of the swing, (including the case $b=0$) and not its suddenness $a$.
Comparison with equation (29) shows some similarity, but many comparable functions have Fourier transforms with a frequency limit at $b$ rather than $a$. A hint about evaluating the integral might be what I need.
 A: Suppose that $a>0$ and $b$ are real numbers. Then the value of this Fourier cosine transform at a real $y$ is
$$\sqrt{\frac2\pi}\int_0^\infty dx\,\cos(xy)\frac{\cos⁡(b\sqrt{a^2+x^2})}{(a^2+x^2)^{3/2}} \\ 
=2\sqrt{\frac2\pi}\int_a^\infty dt\,\cos\big(y\sqrt{t^2-a^2}\,\big)\frac{\cos⁡(bt)}{t^{3/2}}\to0$$
as $b\to\infty$, by the Riemann--Lebesgue lemma. Also, the value of this Fourier cosine transform at $b=0$ is ${\sqrt{\frac{2}{\pi }}}\frac y{a}\, K_1(a y)>0$ if $y>0$. So, this Fourier cosine transform must depend on $b$.
Mathematica cannot find this Fourier cosine transform, which therefore seems unlikely to exist in closed form:

A: According to the Table Errata reported in Mathematics of Computation, vol. 65, no. 215, 1996, pp. 1379–1386, this entry in Erdélyi's Tables of Integral Transforms is flawed. The exponent in the denominator should be $1$ instead of $3/2$, and the condition should be $y>b>0$ instead of $y>a$. Unfortunately, therefore, this table entry does not actually address the Fourier transform required by the OP.
A: For what it is worth, the following related integral
$$
  C(y;b,c) := \int_0^\infty \frac{\cos(c\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}
    \cos(bx) \, dx =
  \begin{cases}
    K_0(y \sqrt{b^2-c^2}) & [b>c>0; y>0] \\
    -\frac{\pi}{2} Y_0(y \sqrt{c^2-b^2}) & [c>b>0; y>0]
  \end{cases} .
$$
appears as formula 2.5.25.15 in

Prudnikov, A. P.; Brychkov, Yu. A.; Marichev, O. I., Integrals and series. Vol. 1. Elementary functions, Moscow: Fiziko-Matematicheskaya Literatura (ISBN 5-9221-0323-7). 632 p. (2003). ZBL1103.00301.

A quick numerical check with Mathematica suggests that the formula checks out. Twice integrating $-C(y;b,c)$ should give your desired cosine transform, up to fixing integration constants.
A once-integrated $C(y;b,c)$ appears in the cited evaluation of another related cosine transform in this answer, which was helpfully linked by MathOverflow under related questions.
