The integral $$\int_0^{\infty} e^u \ K_{i t}(u) du$$ is the adjoint Kontorovich-Lebedev transform of the increasing exponential function, but unfortunately this integral is divergent because $$e^u \ K_{i t}(u) \sim \sqrt{u}^{-1}, \ u\to\infty. $$ $K_{i t}(u)$ is the Modified Bessel Function of purely imaginary order.
In the context of Fourier transforms it is often possible to regularize a divergent Fourier transform so that the outcome is a well-defined distribution. I would like to know if the same thing is possible here. None of the usual tricks that I know work.
For reference, the Fourier transform $$\int_{-\infty}^{\infty} \lvert x \rvert^{-1/2} e^{-i k x} dx = \sqrt{\frac{2 \pi}{k}},$$ according to my table. I do not know how to arrive at this result myself. I can arrive at this result myself by tempering the original function with a Gaussian and then taking the limit that the Gaussian approaches one everywhere. The same trick does not seem to work with my original integral.