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As in the title, I would like to compute the integral:

\begin{equation} \int_{-1}^{1}e^{-1/(1-x^2)}dx \end{equation}

My hunch tells me that I should try to transform it to the correspoding $\int_{-1}^{1} e^{-y^2}dy$ but I am not sure.

Thank you for your time!

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    $\begingroup$ Do you want to approximate it? Because I don't think there's a nice closed form for that integral $\endgroup$ – Gregory Grant May 17 '15 at 11:01
  • $\begingroup$ @Gregory Grant Oh, yes I meant an approximation. The only thing that comes in mind is the Mean Value Theorem of Integral Calculus.. $\endgroup$ – 010514 May 17 '15 at 11:08
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$$\begin{eqnarray*}\color{red}{I}=\int_{-1}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx &=& 2\int_{0}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx=\int_{0}^{1}\exp\left(\frac{1}{z-1}\right)\frac{dz}{\sqrt{z}}\\&=&\int_{0}^{1}\exp\left(-\frac{1}{z}\right)\frac{dz}{\sqrt{1-z}}=\int_{1}^{+\infty}\frac{dt}{t e^t\sqrt{t^2-t}}\\&=&\int_{0}^{+\infty}\frac{e^{-(u+1)}du}{(u+1)\sqrt{u(u+1)}}=\frac{2}{e}\int_{0}^{+\infty}\frac{e^{-\eta^2}\,d\eta}{(1+\eta^2)^{3/2}}\\&=&\color{red}{\frac{\sqrt{\pi}}{e}\,U\left(\frac{1}{2},0,1\right)}\end{eqnarray*}$$

where $U(a,b,z)$ is the Tricomi's confluent hypergeometric function.

If we take the last integral and switch to Fourier transforms, that can be written also as: $$\frac{2}{e\sqrt{\pi}}\int_{0}^{+\infty} e^{-s^2/4} s\, K_1(s)\,ds $$ where $K_1$ is a modified Bessel function of the second kind. Tight numerical approximations follows from the fact that the last integrand function is smooth and essentially supported on $[0,4]$, since the integral over $[4,+\infty)$ is extremely small. We also have: $$ \int_{-1}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = \frac{2}{e}\int_{0}^{\pi/2}e^{-\tan^2 t}\cos t\,dt.$$ Trivial inequalities are: $$\color{red}{0.412\ldots}=\sqrt{\frac{2\pi}{5e^2}}=\frac{2}{e}\int_{0}^{+\infty}e^{-5u^2/2}\,du \leq \color{red}{I}\leq \frac{2}{e}\int_{0}^{+\infty}\frac{du}{(1+u^2)^{5/2}}=\frac{4}{3e} =\color{red}{0.490\ldots}$$

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  • $\begingroup$ Very nice, I've never seen this and the CAS I have access does not calculate it (although it knows about the confluent hypergeometric functions). $\endgroup$ – mickep May 17 '15 at 11:17
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    $\begingroup$ @Jack D'Aurizio Really amazing! Thank you! $\endgroup$ – 010514 May 17 '15 at 11:29
  • $\begingroup$ (+1) nice work @Jack. do you think there is some further simplification of the hypergeometric? The arguments look quiet "easy", but i'm far from being an expert in this kind of manipulations.\ $\endgroup$ – tired May 17 '15 at 12:59
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$\int_{-1}^1e^{-\frac{1}{1-x^2}}~dx$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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  • $\begingroup$ An also nice approach, I had not seen before the hyperbolic substitution with this integral! $\endgroup$ – 010514 May 20 '15 at 22:51

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