Kontorovich–Lebedev transform

Remarks

The integral converges absolutely because $\abs{K_{iy}(x)} \le \int_0^\infty e^{-x\cosh t}\, dt = K_0(x)$. The function $K_{iy}(x)$ is real and even in $y$, so $g$ is even and is usually considered on $(0, \infty)$. Under suitable conditions on $f$, the inversion formula is $$f(x) = \frac{2}{\pi^2 x} \int_0^\infty g(y)\, K_{iy}(x)\, \sinh(\pi y)\, y\, dy.$$

Example: $f(x) = 1$. Since $K_0(x)$ behaves like $-\log x$ near $0$ and like $\sqrt{\pi/(2x)}\, e^{-x}$ at infinity, $K_0$ is integrable and the condition holds. The formula $\int_0^\infty K_\mu(x)\, dx = \frac{\pi}{2\cos(\pi\mu/2)}$ for $\abs{\operatorname{Re} \mu} < 1$, with $\mu = iy$, gives $g(y) = \frac{\pi}{2\cosh(\pi y / 2)}$.

Non-example: $f(x) = e^x$. For fixed $y$, $K_{iy}(x) \sim \sqrt{\pi/(2x)}\, e^{-x}$ as $x \to \infty$, so $f(x) K_{iy}(x) \sim \sqrt{\pi/(2x)}$ and the integral diverges at infinity.