How to evaluate $\int_{0}^{\infty} e^{ - a x } \sinh(\pi x) K_{ix}(b) K_{ix}(c) dx$? I'm interested in evaluating the integral  $\int_{0}^{\infty} e^{ - a x } \sinh( \pi x ) K_{ix}(b) K_{ix}(c) \,dx$, where $0 < a < \pi$, and $b, c >0$. I've notice that if I evaluate the following integral then I should be in the clear:
$$
I(A,B,C) \ = \ \int_{0}^{\infty} e^{ - A x } K_{ix}(B) K_{ix}(C)\,dx
$$ 
Interestingly, in Gradshteyn and Ryzhik (Eq 3 in Ch 6.79) I have encountered the following integral:
$$
\int_{-\infty}^{\infty} e^{(\pi - \gamma)x} K_{ix+iy}(a) K_{ix+iz}(b) = \pi e^{- \beta y - \alpha z} K_{iy-iz}(c)\,dx
$$
where $0<\gamma<\pi$, and $\alpha, \beta,\gamma$ are the angles of the triangle with sides $a,b,c>0$. This reduces to the following integral more similar to mine:
$$
\int_{-\infty}^{\infty} e^{(\pi - \gamma)x} K_{ix}(a) K_{ix}(b)\,dx = \pi K_{0}\left(\sqrt{ a^2 + b^2 - 2 a b \cos( \gamma ) } \right)
$$
This is almost what I need! I just need the limits of integration to match up with mine - I have wasted many hours last night trying to transform this and other integrals in G&R to try and come up with something. Is there any way to do this?
P.S. $K_{ix}(\alpha)$ is of course the modified Bessel function of second kind, of order $ix$, evaluated at the point $\alpha$.
 A: I have looked through various tables of Kontorovich-Lebedev transfroms and found nothing (any lists I have found are quite short).
My attempt at my integral of interest:
$$
G(a,b,c) = \int_{0}^{\infty} dx\ e^{-ax} \sinh( \pi x ) K_{ix}(b) K_{ix}(c) 
$$
On page 111 of Yakubovich's ``the hypergeometric approach to integral transforms and convolutions'' I have found the following identity:
$$
\sinh( \pi x ) K_{ix}(b) K_{ix}(c) = \frac{\pi}{2} \int_{\big| \ln\left( \frac{b}{c} \right) \big|}^{\infty} J_{0} \left( \sqrt{ 2 b c \cosh(u) - b^2 - c^2 } \right) \sin( u x ) du
$$
Which allows me to write:
$$
G(a,b,c) = \frac{\pi}{2} \int_{\big| \ln\left( \frac{b}{c} \right) \big|}^{\infty} du\ J_{0} \left( \sqrt{ 2 b c \cosh(u) - b^2 - c^2 } \right) \frac{u}{u^2 + a^2} 
$$
Doing a coordinate transformation, I was able to get this into the form:
$$
G(a,b,c) = \frac{\pi}{2} \int_{0}^{\infty} d\lambda \ \frac{J_{0}(2\sqrt{\lambda})}{\sqrt{ \left( \lambda + \frac{1}{4}(b-c)^2 \right)\left( \lambda + \frac{1}{4}(b+c)^2 \right) }} \frac{\cosh^{-1}\left( \frac{4 \lambda + b^2 + c^2}{2 b c} \right)}{ \left[ \cosh^{-1}\left( \frac{4 \lambda + b^2 + c^2}{2 b c} \right) \right]^2 + a^2}
$$
My hope in doing this was to take the series expansion of $J_{0}(2\sqrt{\lambda})$, but this has not been fruitful.
Does anyone have any ideas on how to complete this integral (or through a different avenue)?
A: I’ll post what I have so far, as requested and because someone might pick up a way forward from the end result.  It is also too long for a comment but should not be considered a solution to the problem.  I feel that there might actually be a closed from worth chasing.
Firstly the function of interest has the integral representation of
$$u\left( x,y,t \right)=\int\limits_{0}^{\infty }{{{e}^{-ts}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}.$$
It should be noted that this function appears to satisfy a diffusion like differential equation, so it might help to post some background information regarding where all this has come from.  The reason I say that is that these sort of integrals – in terms of Lebedev’s transform / index transforms – have a very natural setting in eigenfunction expansions (see Titchmarsh - eigenfunction expansions vol 1).  So there might be a natural interpretation of it, and it’s properties, within that framework. 
Note the identity 
$${{e}^{-ts}}=\frac{2}{\pi }\int\limits_{0}^{\infty }{\frac{w}{{{w}^{2}}+{{s}^{2}}}\sin \left( wt \right)du}\ \ \ \ \operatorname{Re}\left( t,s \right)>0$$
And so we have 
$$\frac{\partial }{\partial t}u\left( x,t \right)=-\int\limits_{0}^{\infty }{\frac{s}{{{w}^{2}}+{{s}^{2}}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}\frac{2}{\pi }\int\limits_{0}^{\infty }{w\sin \left( wt \right)dw}$$
Now from Bateman’s Integral Transforms vol. 2 pg 176 12.1.8 we have
$$\int\limits_{0}^{\infty }{\frac{s}{{{n}^{2}}+{{s}^{2}}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}=\left\{ \begin{matrix}
   \frac{{{\pi }^{2}}}{2}{{I}_{n}}\left( y \right){{K}_{n}}\left( x \right) & 0<y<x  \\
   \frac{{{\pi }^{2}}}{2}{{I}_{n}}\left( x \right){{K}_{n}}\left( y \right) & x<y<\infty   \\
\end{matrix} \right.$$
This is for n, a natural number.  Suppose we can analytically continue it then we would have for example
$$\frac{\partial }{\partial t}u\left( x,y,t \right)=-\pi \int\limits_{0}^{\infty }{w{{I}_{w}}\left( y \right){{K}_{w}}\left( x \right)\sin \left( wt \right)dw}$$
for $0<y<x$, and a similar representation for $x<y$ etc.  Integrating we find 
$$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{I}_{w}}\left( y \right){{K}_{w}}\left( x \right)\cos \left( wt \right)dw}+{{c}_{1}}\left( x,y \right)\ \ \ \ 0<y<x$$ 
$$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{I}_{w}}\left( x \right){{K}_{w}}\left( y \right)\cos \left( wt \right)dw}+{{c}_{2}}\left( x,y \right)\ \ \ \ 0<x<y$$
Note:
${{K}_{w}}\left( x \right)\sim \sqrt{\frac{\pi }{2w}}{{\left( \frac{ex}{2w} \right)}^{-w}}$ and ${{I}_{w}}\left( x \right)\sim \sqrt{\frac{\pi }{2w}}{{\left( \frac{ex}{2w} \right)}^{w}}$  for large w.  Now observe that from the original representation that $\underset{t\to \infty }{\mathop{\lim }}\,u\left( x,y,t \right)=0$, by the Riemann-Lebesgue lemma.  Therefore also by RL the first term in the above representations becomes zero in the limit, and hence we must have therefore ${{c}_{1}}={{c}_{2}}=0$.  We have then
$$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{\kappa }_{w}}\left( x,y \right)\cos \left( wt \right)dw}$$
where $\kappa$ takes on the different forms depending on $x<y$ or $y<x$.  At the moment the reason why I think all this works and is worth pursuing is that a numerical integration comparison of this representation to that of the original representation is spot-on to many decimal places for a wide range of x,y that I have tested.  Obviously not a proof, but I think one can ‘feel' that the justification to move n away from the naturals and to a real number wouldn’t be a stretch.  Another reason is that if you now use Lebedev’s formula (pg 140 from special functions), you will arrive at the formula you have already derived (second last formula).  Since Lebedev's formula is valid for arbitrary indices, then this might end up constituting a 'proof' of continuation.
Note from the form above we have
$${{\hat{u}}_{C}}\left( x,y,w \right)=\pi {{K}_{w}}\left( x \right){{I}_{w}}\left( y \right)$$
where ${{\hat{u}}_{C}}$ is the cosine transform of u, which may be useful (?).  I have various other representations but nothing that is heading to anything I’d call ‘closed’.  
