$$\int_{0}^{1} e^{\frac{1}{\log(x)}} \, dx = 2 \,K_{1}(2)$$,
where $K_1$ is the modified Bessel function of the second kind. I was wondering,
$(1)$ Can you express $$\int_{0}^{1} e^{\frac{1}{\log(x)}}*e^{\frac{1}{\log(1-x)}} \, dx$$ in terms of a Bessel function?
$(2)$ Why does $$\int_{0}^{\infty} x^3 \, K_{1}(x^2) \, dx = \frac{\pi}{4} \, ?$$
$(3)$ Why does $$\int_{0}^{\infty} \sqrt{x} \, K_{1}(\sqrt{x}) \, dx = 4 \, ? $$
For the first integral, the substitution $x = e^{-1/\xi}$ gives the Mellin transform of $e^{-\xi - 1/\xi}$: $$ \mathcal M[e^{-\xi - 1/\xi}] = 2 K_{-s}(2), \\ \int_0^1 e^{1 / \ln x} dx = \mathcal M[e^{-\xi - 1/\xi}](-1).$$ $e^{1 / \ln x} e^{1 / \ln(1 - x)}$ does not seem to be related to this transform (I'm assuming $*$ means a product).
For the second question, the Mellin transform of $K_1$ is $$\mathcal M[K_1] = 2^{s - 2} \Gamma \!\left( \frac {s - 1} 2 \right) \Gamma \!\left( \frac {s + 1} 2 \right), \\ \int_0^\infty x^p K_1(x^q) dx = \frac 1 {|q|} \mathcal M[K_1] \!\left( \frac {p + 1} q \right), \quad \frac {p + 1} q > 1.$$ When $(p + 1)/q$ is integral, we get gamma functions of integral or half-integral arguments.