Is this related to any integral transform?
$$\int_0^1 \Psi(x)\Psi(1-x)\,dx=\int_{0}^{1} e^{{\frac{1}{\log(x)}}+{\frac{1}{\log(1-x)}}} \, dx.$$
The integral, where $K$ is the modified Bessel function of the second kind, $$ \int_{0}^{1} e^{{\frac{1}{\log(x)}}} \, dx =2K_1(2), $$
Can be evaluated using the substitution $x = e^{-1/\xi},$ which gives the Mellin transform of $e^{-\xi - 1/\xi}:$
$$\mathcal M[e^{-\xi - 1/\xi}] = 2 K_{-s}(2), \\ \int_0^1 e^{1 / \ln x} dx = \mathcal M[e^{-\xi - 1/\xi}](-1).$$
I think the Mellin transform is also related to: $$ \int_{0}^{1} e^{{\frac{1}{\log(1-x)}}} \, dx. $$
However $$ \int_0^1 \Psi(x)\Psi(1-x) \,dx $$
does not seem to be related to the Mellin transform.