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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.

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Did you consider the error function? en.wikipedia.org/wiki/Error_function –  Bernhard May 14 '12 at 14:07
    
@Bernhard: it doesn't apply here, surprisingly enough. –  J. M. May 14 '12 at 14:23
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Mathematica gives a closed form (after a fair bit of coaxing) in terms of Meijer's $G$-function: $$\sqrt{\pi}\,\mathrm{G}_{1,2}^{2,0}\left(1\mid{{\frac32}\atop{0,1}}\right)$$ –  J. M. May 14 '12 at 14:47

3 Answers 3

up vote 3 down vote accepted

I have the best approach:

$\int_{-1}^1e^{\frac{1}{x^2-1}}~dx$

$=\int_{-1}^0e^{\frac{1}{x^2-1}}~dx+\int_0^1e^{\frac{1}{x^2-1}}~dx$

$=\int_1^0e^{\frac{1}{(-x)^2-1}}~d(-x)+\int_0^1e^{\frac{1}{x^2-1}}~dx$

$=\int_0^1e^{\frac{1}{x^2-1}}~dx+\int_0^1e^{\frac{1}{x^2-1}}~dx$

$=2\int_0^1e^{\frac{1}{x^2-1}}~dx$

$=2\int_0^\infty e^{\frac{1}{\tanh^2x-1}}~d(\tanh x)$

$=2\int_0^\infty e^{-\frac{1}{\text{sech}^2x}}~d(\tanh x)$

$=2\int_0^\infty e^{-\cosh^2x}~d(\tanh x)$

$=2\left[e^{-\cosh^2x}\tanh x\right]_0^\infty-2\int_0^\infty\tanh x~d\left(e^{-\cosh^2x}\right)$

$=4\int_0^\infty e^{-\cosh^2x}\sinh x\cosh x\tanh x~dx$

$=4\int_0^\infty e^{-\cosh^2x}\sinh^2x~dx$

$=4\int_0^\infty e^{-\frac{\cosh2x+1}{2}}\dfrac{\cosh2x-1}{2}dx$

$=2e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh2x}{2}}(\cosh2x-1)~dx$

$=e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh2x}{2}}(\cosh2x-1)~d(2x)$

$=e^{-\frac{1}{2}}\int_0^\infty e^{-\frac{\cosh x}{2}}(\cosh x-1)~dx$

$=e^{-\frac{1}{2}}\left(K_1\left(\dfrac{1}{2}\right)-K_0\left(\dfrac{1}{2}\right)\right)$

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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.

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Note that the Whittaker function $W$ is easily recast as a Tricomi confluent hypergeometric function $U$ (which is probably more likely to be implemented in a computing environment than the Whittaker functions): $$\frac{\sqrt\pi}{e}\,U\left(\frac12,0,1\right)$$ –  J. M. May 15 '12 at 1:07
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The Tricomi function, then, easily degenerates to a difference of modified Bessel functions of the second kind: $$\frac{K_1(1/2)-K_0(1/2)}{\sqrt e}$$ I suppose you can't get more elementary than that... –  J. M. May 15 '12 at 1:13
    
@J.M. Ah, it was so close. Great answer in the end! –  Aleksey Pichugin May 15 '12 at 7:36

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) $.

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