The following answer shows that $$\small \left[ K_{ib}(a) \right]^{2} + \frac{\pi}{2} \, \cosh(\pi b) \int_{0}^{\infty} Y_{0}\left(2 a \sinh \frac{t}{2} \right) \cos(bt) \, \mathrm dt + \frac{\pi}{2} \, \sinh(\pi b) \int_{0}^{\infty} J_{0}\left(2 a \sinh \frac{t}{2} \right) \sin(bt) \, \mathrm dt =0. $$
The two integral identities satisfy this equation since $- \cosh^{2}(\pi b)+ \sinh^{2}(\pi b) = -1$.
This equation can be used with Gary's answer to prove the second integral identity.
My starting point is the more well known integral representation $$\left[ K_{\nu}(x) \right]^{2} = 2 \int_{0}^{\infty}K_{0}(2x \cosh t) \cosh(2 \nu t) \, \mathrm dt , \, \quad x>0. $$
Using this representation, we have $$\left[ K_{ib}(a) \right]^{2} = 2 \int_{0}^{\infty} K_{0} (2a \cosh x) \cos(2bx) \, \mathrm dx = \int_{0}^{\infty} K_{0} \left(2a \cosh \frac{u}{2} \right) \cos(bu) \, \mathrm du.$$
The important thing to notice here is that $\left[ K_{ib}(a) \right]^{2}$ is a real value if $a$ and $b$ are positive.
The function $$ f(z) = K_{0} \left(2a \cosh \frac{z}{2} \right) \cos(bz) $$ has branch cuts were $\cosh \left(\frac{x}{2} \right)$ is real and negative.
None of these branch cuts fall inside or on a rectangular contour with vertices at $z=0$, $z= R$, $z= R +\pi i $, and $z= \pi i $.
There is, however, a branch point on the contour at $z= i \pi$.
So let's integrate $f(z)$ counterclockwise around the above contour with the addition of an quarter circle indentation about $z = i \pi$.
However, since $ \lim_{z \to i \pi} (z- i \pi)K_{0} \left(2a \cosh \frac{z}{2} \right) \cos(bz) =0$, there is no contribution from letting the radius of the indentation go to zero. (The function $K_{0}\left( 2a \cosh \frac{z}{2} \right)$ behaves like $-\log(z-i \pi)$ near $z= i \pi$.)
As $R \to \infty$, the integral vanishes on the right side of contour because the magnitude of $K_{0}(z)$ decays exponentially to zero as $\Re(z) \to + \infty$.
On the left side of the contour, we have $$-i \int_{0}^{\pi} K_{0} \left(2a \cos \frac{t}{2} \right) \cosh (bt) \, \mathrm dt. $$
And on the top side of the contour, we have $$ \begin{align} &-\int_{0}^{\infty} K_{0} \left (2a \cosh \frac{t+ i \pi}{2} \right) \cos \left(b(t + i \pi) \right) \, \mathrm dt \\ &= - \int_{0}^{\infty} K_{0} \left(2ai \sinh \frac{t}{2} \right) \left(\cosh(\pi b)\cos(bt) - i \sinh(\pi b) \sin(bt) \right) \, \mathrm dt \\ & \overset{(1)}{=} \frac{\pi}{2} \int_{0}^{\infty}\left(Y_{0} \left(2a \sinh \frac{t}{2} \right) + i J_{0} \left(2a \sinh \frac{t}{2} \right) \right)\left(\cosh(\pi b)\cos(bt) - i \sinh(\pi b) \sin(bt) \right) \, \mathrm dt. \end{align}$$
Therefore, since there are no singularities inside the contour, we have $$\left[ K_{ib}(a) \right]^{2} + \frac{\pi}{2} \int_{0}^{\infty}\left(Y_{0} \left(2a \sinh \frac{t}{2} \right) + i J_{0} \left(2a \sinh \frac{t}{2} \right) \right)\left(\cosh(\pi b)\cos(bt) - i \sinh(\pi b) \sin(bt) \right) \, \mathrm dt$$
$$-i \int_{0}^{\pi} K_{0} \left(2a \cos \frac{t}{2} \right) \cosh (bt) \, \mathrm dt =0. $$
And by equating the real parts on both sides of the above equation, we get $$ \small \left[ K_{ib}(a) \right]^{2} + \frac{\pi}{2} \, \cosh(\pi b) \int_{0}^{\infty} Y_{0}\left(2 a \sinh \frac{t}{2} \right) \cos(bt) \, \mathrm dt + \frac{\pi}{2} \, \sinh(\pi b) \int_{0}^{\infty} J_{0}\left(2 a \sinh \frac{t}{2} \right) \sin(bt) \, \mathrm dt =0. $$
$(1)$ See the answer here.
Best Answer
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.