The objective is to:
Evaluate $$ \int_0^1 e^{\frac{1}{\log(\theta)}} ~d\theta $$ in polar coordinates.
Using cartesian coordinates:
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 $e^{-\xi - 1/\xi}:$
$$\mathcal M[e^{-\xi - 1/\xi}] = 2 K_{-s}(2), \\ \int_0^1 e^{1 / \ln x} dx = \mathcal M[e^{-\xi - 1/\xi}](-1).$$
$$ I=\int_{0}^{1} e^{1/\ln \theta} d\theta$$ Let $1/\ln theta=x$ then $$I=-\int_{0}^{-\infty} e^{x+1/x}~ \frac{1}{x^2} dx= \int_{0}^{\infty} e^{-(y+1/y)} \frac{1}{y^2} dy$$ Let $y=e^{u}$ then $$I=\int_{-\infty}^{\infty} e^{-u} e^{-(e^u+e^{-u})} du = \int_{-\infty}^{\infty} (\cosh u -\sinh u) e^{-2\cosh u} du$$ $$\implies I=2 \int_{0}^{\infty} \cosh u~ e^{-2\cosh u} du$$ We have used the evenness and oddness of the integrand. Using the standard defibition of cylinderical modifies Bessel function of second kinf ($K_{\nu}(z))$ we get $$I=2 K_1(2)$$ See https://en.wikipedia.org/wiki/Bessel_function