$\int_{-1}^{1}e^{-\frac{1}{1-x^2}}dx$, can it be computed? Is there a way to compute $\int^{1}_{-1} e^{-\frac{1}{1-x^2}}dx$ ?
I have tried a few change of variables and also to write down $\frac{1}{1-x^2} = \frac{1}{2} ( \frac{1}{1-x} + \frac{1}{1+x})$ But I didn't get anything so far.
Edit: changed $e^{\frac{1}{1-x^2}}$ to $e^{-\frac{1}{1-x^2}}$

$$\int_{-1}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = 2\int_{0}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = \int_{0}^{1}\exp\left(\frac{1}{x-1}\right)\frac{dx}{\sqrt{x}}$$
equals:
$$ \int_{0}^{1}\exp\left(-\frac{1}{x}\right)\frac{dx}{\sqrt{1-x}} = \int_{1}^{+\infty}\frac{e^{-x}}{x^{3/2}\sqrt{x-1}}\,dx=\frac{1}{e}\int_{0}^{+\infty}\frac{e^{-x}\,dx}{(x+1)^{3/2}\sqrt{x}} $$
or:
$$ \frac{2}{e}\int_{0}^{+\infty}\frac{e^{-x^2}}{(x^2+1)^{3/2}}\,dx = \frac{\sqrt{\pi}}{e}\,U\left(\frac{1}{2},0,1\right)\approx 0.443993816168$$
where $U$ is Tricomi's confluent hypergeometric function. An accurate and simple upper bound is given by the Cauchy-Schwarz inequality:
$$ \int_{0}^{+\infty}\frac{e^{-x^2}}{(x^2+1)^{3/2}}\,dx \leq \sqrt{\int_{0}^{+\infty}e^{-2x^2}\,dx\int_{0}^{+\infty}\frac{dx}{(x^2+1)^3}}$$
gives that the original integral is $\leq \frac{\sqrt{3}}{2e}\cdot\left(\frac{\pi}{2}\right)^{3/4}$.
