# Is $\int_0^1 \Psi(x)\Psi(1-x)\,dx$ related to any transform?

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.

• 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 ? – reuns Mar 9 at 0:48
• So it is related to the Mellin transform? Is there a closed form for the value of the integral? – Ultradark Mar 10 at 18:21