Need your help with the integral $\int_0^\infty\frac{dx}{e^{\,e^{-x}} \cdot e^{\,e^{x}}}$.

Question

Is it possible to evaluate this integral in a closed form? $$\int_0^\infty\frac{dx}{e^{\,e^{-x}} \cdot e^{\,e^{x}}}$$

Answer 1

This is clearly equivalent to $$\int_0^{\infty}e^{-2\cosh x}dx=K_0(2).$$ Here $K_0(x)$ denotes Macdonald function, which has integral representation $$K_{\nu}(r)=\int_0^{\infty}e^{-r\cosh x}\cosh\nu x\,dx.$$