For any $\rho\in\mathbb{R}^+$, prove that the following formula equals a constant:
$$\dfrac{1}{\rho^2}{\int_{-\rho}^\rho x^2 e^{\left(\tfrac{\rho^2}{x^2-\rho^2}\right)}dx}\left({\int_{-\rho}^\rho e^{\left(\tfrac{\rho^2}{x^2-\rho^2}\right)}dx}\right)^{-1}=\text{constant}$$
The constant can be easily verified as about 0.158..., but how to prove it?
Can we obtain a radical form of the constant? Is it a transcendental number?
Let's change variables in the integral: $x=\rho y$, $dx = \rho dy$: $$ \dfrac{1}{\rho^2}{\int_{-\rho}^\rho x^2e^\tfrac{\rho^2}{x^2-\rho^2}dx}\left({\int_{-\rho}^\rho e^\tfrac{\rho^2}{x^2-\rho^2}dx}\right)^{-1}= \dfrac{1}{\rho^2}{\int_{-1}^1 \rho^2 y^2e^\tfrac{1}{y^2-1}\rho dy}\left({\int_{-1}^1 e^\tfrac{1}{y^2-1}\rho dy}\right)^{-1}= {\int_{-1}^1 e^\tfrac{1}{y^2-1} dy}\left({\int_{-1}^1 y^2 e^\tfrac{1}{y^2-1} dy}\right)^{-1} $$
added: However it is apparent that the integral is not convergent.