The first thing I have to note is that I am not 100% sure whether this problem is due to Mathematica or just the mathematics I have produced. Thus feel free to direct me to the appropriate forum.
Let $f(x) = e^{\frac{-1}{1 - x^2}}$, $g(x) = \frac{d^2f(x)}{dx^2}$. Then $g(x)^2$ has a finite area as is confirmed by the picture below:
Due to a specific example I am working with I need to use change of variables $u = \frac{\pi}{4}x$ so that $f$'s domain would be $\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)$ and $f(u) = e^{\frac{-1}{1 - \left(\frac{4x}{\pi}\right)^2}}$.
As I am not that savy with Mathematica I reckoned that it would take me less time to compute everything with $f(u)$ than with $f(x)$ and somehow have Mathematica to perform the change of variables.
What I end up with is (where I am using the symbol $x$ instead of $u$)
which isn't at all what I wanted. Is there some subtlety I am overlooking why else wouldn't the integral evaluate after the change of variables?
I did not have any problem $$f(x)=e^{-\frac{1}{1-x^2}}\quad \implies \quad g(x)=f''(x)=\frac{2(3 x^4-1)}{\left(x^2-1\right)^4}e^{-\frac{1}{1-x^2}}$$ and numerical integration gives $$\int_{-\frac \pi 4}^{+\frac \pi 4} \Bigg[\frac{2(3 x^4-1)}{\left(x^2-1\right)^4}e^{-\frac{1}{1-x^2}} \Bigg]^2\,dx=1.8340764522074691939818132792902350094\cdots$$
Without any specific option (except the
WorkingPrecision, I did not have any problem to get $500^+$ decimal places.Even if it does not mean anything, a good approximation (absolute error of $2.9\times 10^{-7}$) is $$\frac{150 \left(1-2 \tan ^{-1}\left(\frac{1}{2}\right)\right)}{1+3 \sqrt{e}}$$