# Integral representation of Bessel function K

Does someone have an idea how to connect the following function (appearing in the quantization of a real scalar field in a uniformely accelerated frame) : $$K(x,y) = \int_{0}^{\infty} \frac{dt}{t} t^{ix} exp\left(iy\left[\frac{1}{t}-t\right]\right)$$ ($x$ and $y$ are real) to the standard Bessel functions ? Actually it doesn't even seem to me that this integral is well defined (properly convergent)... The integrand looks like the one for Schläfli’s contour integral representation of $J_v(x)$ : $$J_{\nu}(z) = \frac{1}{2\pi i}\int_C dt \frac{1}{t^{\nu + 1}} exp\left(\frac{z}{2}\left[t-\frac{1}{t}\right]\right)$$ where $C$ encircles the origin, but I don't see how the contour could be deformed. Moreover $K$ satisfies also to a (modified) Bessel equation (however I could not kill the boundary terms), so I should be able to relate them...

Thank you a lot if you can help!

The integral can be seen to be the Mellin transform of $e^{iy(1/t-t)}$ $$I=\int_0^\infty t^{s-1}e^{iy(1/t-t)}\;dt \tag{1}$$ with $s=ix$. The result of $(1)$ is $$I=2 K_s(2|y|)\left(\cos\left(\frac{\pi s}{2}\right)-i \;\text{sgn}(y)\sin\left(\frac{\pi s}{2}\right)\right)$$ so then you have $$I=2 K_{ix}(2|y|)\left(\cosh\left(\frac{\pi s}{2}\right)-\;\text{sgn}(y)\sinh\left(\frac{\pi s}{2}\right)\right)$$ I'm not sure how one would interpret the complex index of the Bessel $K$ function. Perhaps via the integral representation $$K_\nu(z) = \int_0^\infty e^{-z \cosh(t)}\cosh(\nu t)\;dt$$ such that $\cosh \to \cos$ and it looks like half a Fourier cosine transform. Seems to reproduce numerical tests for a few $x$ and $y$.