Whittaker transform

Remarks

Other authors put other powers of $xy$ and other exponential factors in front of $W_{\kappa, \mu}(xy)$; the name covers this whole family of kernels.

Example: for $\kappa = \mu + 1/2$ one has $W_{\mu + 1/2, \mu}(z) = e^{-z/2} z^{\mu + 1/2}$. Then $k(z) = e^{-z/2} z^{-\mu - 1/2} e^{-z/2} z^{\mu + 1/2} = e^{-z}$, and the transform is the Laplace transform $g(y) = \int_0^\infty e^{-xy} f(x)\, dx$.

Non-example: $W_{\kappa, \mu}(z)$ behaves like $e^{-z/2} z^{\kappa}$ as $z \to \infty$, so $k(xy)$ behaves like $e^{-xy}$ for large $x$. Hence $f(x) = e^{x^2}$ gives a divergent integral for every $y > 0$ and has no Whittaker transform.