Proof of integral Is there an analytical method to show that
$$
\int_{-a}^a\exp\left(\frac{-1}{1-(x/a)^2}\right)\,\mathrm{d}x=ka,
$$
for $a>0$.
I have confirmed this result numerically for a range of values of $a$. This numerical investigation gave $k\approx0.439938161681\pm5\times10^{-13}$.
This integral arose while attempting to approximate a function. I originally planned to just evaluate this integral numerically until I came across the above relationship. 
Applying Leibniz rule for differentiation w.r.t. $a$ under the integral didn't lead to any obvious simplifications.
Is there a technique that can be used to confirm this relationship? Or is it just a coincidence?
 A: If you are just looking to confirm the relationship, then the problem is not difficult.  If, however, you are looking for a closed, exact, form for $k$, I don't know what to tell you.
First observe that
$$
\int_{-1}^{1} \exp\left( \frac{-1}{1 - x^2} \right) dx = k
$$
is well-defined (ie: the integrand is integrable).  Now do a change of variables, sending $x$ to $x/a$ to get:
$$
\int_{-a}^{a} \exp\left( \frac{-1}{1 - (x/a)^2} \right) \frac{dx}{a} = k.
$$
This is the relationship you have above (after bringing $a$ to the RHS of course).  So this relationship is by no means a coincidence, it's follows from that fact that the integral is just being scalled by a factor of $a$ and hence the answer is being scaled by a factor of $a$.
A: Use parameter substitution to solve this:
$$
\text{Let }u = \frac{x}{a},\text{ then}\\
\int_{-a}^a\exp\left(\frac{-1}{1-(x/a)^2}\right)\,\mathrm{d}x=a\int_{-1}^1 \exp\left(\frac{-1}{1-u^2}\right)\mathrm{d}u\\
$$
This you can then solve by other methods (I'd try trig substitution), although it is sufficent to observe that $\int_{-1}^1 \exp\left(\frac{-1}{1-u^2}\right)\mathrm{d}u$ is constant WRT $a$.  Wolfram|Alpha gives me 0.443993816..., for it, but the ISC doesn't seem to have any particularily likely looking results, and I suspect that it is non-elementary.
