# Evaluate definite integral $\int_{-1}^1 \exp(1/(x^2-1)) \, dx$

How to review the adhering to precise integral:

$$\int_{-1}^1 \exp\left(\frac1{x^2-1}\right) \, dx$$

It appears that uncertain indispensable additionally can not be shared in typical features. I would certainly such as any kind of remedy in preferred primary or non - primary features.

19
2022-07-25 20:41:51
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Using the proportion $x \to -x$ and also adjustment 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 after that review as $bkslshrm e ^ - 1/2 bkslshleft ( bkslshrm K _ 1bkslshleft (1/2bkslshright) - bkslshrm K _ 0bkslshleft (1/2bkslshright) bkslshright)$.

10
2022-07-25 22:38:06
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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 . 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 and , 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.

13
2022-07-25 21:15:16
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