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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.

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    $\begingroup$ With $x = e^{-y},y=1/z$ then $\int_0^1 e^{1/\log(x)}dx=\int_0^\infty e^{-1/y-y}dy=\int_0^\infty e^{-1/z-z}z^{-2}dz$. You can apply the same substitution in $\int_0^1 e^{1/\log(x)+1/\log(1-x)}dx$ but it won't simplify. What do you expect more ? $\endgroup$ – reuns Mar 9 at 0:48
  • $\begingroup$ So it is related to the Mellin transform? Is there a closed form for the value of the integral? $\endgroup$ – Ultradark Mar 10 at 18:21

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