Evaluate definite integral $\int_{-1}^1 \exp(1/(x^2-1)) \, dx$ How to evaluate the following definite integral:
$$\int_{-1}^1 \exp\left(\frac1{x^2-1}\right) \, dx$$
It seems that indefinite integral also cannot be expressed in standard functions. I would like any solution in popular elementary or non-elementary functions.
 A: The symmetry of the integrand and variable substitution $t=1/(1-x^2)-1$ can be used to transform your integral as follows:
$$
I=\int_{-1}^1 \exp\left(\frac{1}{x^2-1}\right)dx=\int_0^1 2\exp\left(\frac{1}{x^2-1}\right)dx
=\frac{1}{e}\int_0^{\infty}\frac{e^{-t}dt}{\sqrt{t}(1+t)^{3/2}}\,,
$$
Maple can evaluate this integral in terms of Meijer's $G$-function, just as obtained by J.M. after "coaxing" his Mathematica. Alternatively, this integral can be recognised as Whittaker's function $W$. The same conclusion can also be arrived at by noting that this integral is, effectively, a Laplace transform and using an appropriate command of Maple (or Mathematica), with the result
$$
I=\sqrt{\frac{\pi}{e}} W_{-\frac{1}{2},-\frac{1}{2}}(1) \approx 0.44399\,.
$$
This answer is only slightly more elegant than J.M.'s; it's still not elementary and I am unsure whether you would describe Whittaker's function as "popular". You may also consider reformulating this in terms of confluent hypergeometric function.
Updated 15/05/2012: It seems that you can avoid using Whittaker's function after all, at the cost of computing a modified Bessel function and a certain continued fraction. Specifically, identities 13.17.3 and 13.18.9, given in the "new" DLMF, lead to the simple result:
$$
I=\frac{K_0(1/2)}{C\sqrt{e}}\,,
$$
with constant $C$ given by the following continued fraction:
$$
C=1+\frac{1/2}{1+}\,\frac{3/2}{1+}\,\frac{3/2}{1+}\,\frac{5/2}{1+}\,\frac{5/2}{1+}\,
\frac{7/2}{1+}\,\frac{7/2}{1+}\,\cdots
=\frac{W_{0,0}(1)}{W_{-\frac{1}{2},-\frac{1}{2}}(1)}\approx 1.2628295456\,.
$$
In terms of functions involved, this is a lot more "elementary". From the computational point of view, I suspect that the original answer above is more practical.
A: Using the symmetry $x \to -x$ and  change of variables $t = 1/\sqrt{1-x^2}$ we get
$2 \int_1^\infty \dfrac{e^{-t^2}}{t^2 \sqrt{t^2-1}} dt$, which Maple 16 can then evaluate as
${{\rm e}^{-1/2}} \left( {{\rm K}_1\left(1/2\right)}-
{{\rm K}_0\left(1/2\right)} \right) $.
A: I have the best approach:
$\int_{-1}^1e^{\frac{1}{x^2-1}}~dx$
$=\int_{-1}^0e^{\frac{1}{x^2-1}}~dx+\int_0^1e^{\frac{1}{x^2-1}}~dx$
$=\int_1^0e^{\frac{1}{(-x)^2-1}}~d(-x)+\int_0^1e^{\frac{1}{x^2-1}}~dx$
$=\int_0^1e^{\frac{1}{x^2-1}}~dx+\int_0^1e^{\frac{1}{x^2-1}}~dx$
$=2\int_0^1e^{\frac{1}{x^2-1}}~dx$
$=2\int_0^\infty e^{\frac{1}{\tanh^2x-1}}~d(\tanh x)$
$=2\int_0^\infty e^{-\frac{1}{\text{sech}^2x}}~d(\tanh x)$
$=2\int_0^\infty e^{-\cosh^2x}~d(\tanh x)$
$=2\left[e^{-\cosh^2x}\tanh x\right]_0^\infty-2\int_0^\infty\tanh x~d\left(e^{-\cosh^2x}\right)$
$=4\int_0^\infty e^{-\cosh^2x}\sinh x\cosh x\tanh x~dx$
$=4\int_0^\infty e^{-\cosh^2x}\sinh^2x~dx$
$=4\int_0^\infty e^{-\frac{\cosh2x+1}{2}}\dfrac{\cosh2x-1}{2}dx$
$=2e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh2x}{2}}(\cosh2x-1)~dx$
$=e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh2x}{2}}(\cosh2x-1)~d(2x)$
$=e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh x}{2}}(\cosh x-1)~dx$
$=e^{-\frac{1}{2}}\left(K_1\left(\dfrac{1}{2}\right)-K_0\left(\dfrac{1}{2}\right)\right)$
