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Is there a way to compute $\int^{1}_{-1} e^{-\frac{1}{1-x^2}}dx$ ?

I have tried a few change of variables and also to write down $\frac{1}{1-x^2} = \frac{1}{2} ( \frac{1}{1-x} + \frac{1}{1+x})$ But I didn't get anything so far.

Edit: changed $e^{\frac{1}{1-x^2}}$ to $e^{-\frac{1}{1-x^2}}$

incas
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1 Answers1

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$$\int_{-1}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = 2\int_{0}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = \int_{0}^{1}\exp\left(\frac{1}{x-1}\right)\frac{dx}{\sqrt{x}}$$ equals: $$ \int_{0}^{1}\exp\left(-\frac{1}{x}\right)\frac{dx}{\sqrt{1-x}} = \int_{1}^{+\infty}\frac{e^{-x}}{x^{3/2}\sqrt{x-1}}\,dx=\frac{1}{e}\int_{0}^{+\infty}\frac{e^{-x}\,dx}{(x+1)^{3/2}\sqrt{x}} $$ or: $$ \frac{2}{e}\int_{0}^{+\infty}\frac{e^{-x^2}}{(x^2+1)^{3/2}}\,dx = \frac{\sqrt{\pi}}{e}\,U\left(\frac{1}{2},0,1\right)\approx 0.443993816168$$ where $U$ is Tricomi's confluent hypergeometric function. An accurate and simple upper bound is given by the Cauchy-Schwarz inequality: $$ \int_{0}^{+\infty}\frac{e^{-x^2}}{(x^2+1)^{3/2}}\,dx \leq \sqrt{\int_{0}^{+\infty}e^{-2x^2}\,dx\int_{0}^{+\infty}\frac{dx}{(x^2+1)^3}}$$ gives that the original integral is $\leq \frac{\sqrt{3}}{2e}\cdot\left(\frac{\pi}{2}\right)^{3/4}$.

Jack D'Aurizio
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  • this can be further reduced to bessel functions... http://www.wolframalpha.com/input/?i=tricomiU%5B1%2F2,0,1%5D – tired Feb 16 '17 at 18:55
  • @tired: sure, $K_0$ and $K_1$, arising from the normal ratio distribution, for instance. That is no wonder, since the integrand function is (more or less) given by the product of a Cauchy PDF and a normal PDF. – Jack D'Aurizio Feb 16 '17 at 19:00