Evaluate integral $\int_{-1}^{1} x^2 \exp(\frac{1}{x^2-1}) dx$

Question

How to evaluate this integral?

$\int_{-1}^{1} x^2 \exp(\frac{1}{x^2-1}) dx$

If possible, I want any solution in elementary/non-elementary functions.

Many thanks.

Answer 1

By substituting $\cosh^2t=\dfrac1{1-x^2}$ and exploiting the parity of the integrand, we finally

arrive at $I=2\Big[I_2(1)-I_4(1)\Big]$, where $I_k(a)=\displaystyle\int_0^\infty\frac{e^{-a\cosh^2x}}{\cosh^kx}~dx.~$ Also, we notice that

$-I'_2(1)=I''_4(1)=\displaystyle\int_0^\infty e^{-a\cosh^2x}~dx,~$ where the differentiation happens with regard to

the parameter a. Using the fact that $\cosh^2x=\dfrac{1+\cosh2x}2$ along with our knowledge

of Bessel functions, we evaluate the last integral to equal $\dfrac12~K_0\bigg(\dfrac a2\bigg)~\exp\bigg(\!\!-\dfrac a2\bigg).$

Unfortunately, it would seem that the latter's first and second order anti-derivatives cannot

be expressed in terms of Bessel functions. $($I might be wrong, and I hope I am, but so far

I've had no luck in finding such a closed form for them$)$. However, not all is lost, since we

can always count on the Meijer G-function to come to our rescue. Thus, $$I=\dfrac{\sqrt\pi}2~G_{1,2}^{2,0}\bigg(1;\displaystyle{5/2\choose0,1}\bigg).$$ If someone could rewrite this last expression in terms of Bessel or Struve functions, I would

be very, very grateful.