3
$\begingroup$

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 $e^{-\frac{1}{1-x^2}}$

  • 0
    In that case, then WolframAlpha tells us the answer is [approx 0.443994](https://www.wolframalpha.com/input/?i=integrate+%7Be%5E-%7B1%2F%7B1-x%5E2%7D%7D%7D+from+-1+to+1) – 2017-02-15
  • 0
    Any chance this can be computed analytically to yield an exact value ? – 2017-02-15
  • 0
    Unfortunately I don't have the knowledge to do that, was just drawing your attention to WolframAlpha which I find really useful to see what I am aiming for when I am stuck on a question. If we remove the limits however, it tell us that there is [no standard form for the indefinite integral](https://www.wolframalpha.com/input/?i=integrate+%7Be%5E-%7B1%2F%7B1-x%5E2%7D%7D%7D) which leads me to conclude that you can't find an exact value – 2017-02-15

1 Answers 1

2

$$\int_{-1}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = 2\int_{0}^{1}\exp\left(\frac{1}{x^2-1}\right)\,dx = \int_{0}^{1}\exp\left(\frac{1}{x-1}\right)\frac{dx}{\sqrt{x}}$$ 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 $\leq \frac{\sqrt{3}}{2e}\cdot\left(\frac{\pi}{2}\right)^{3/4}$.

  • 0
    this can be further reduced to bessel functions... http://www.wolframalpha.com/input/?i=tricomiU%5B1%2F2,0,1%5D – 2017-02-16
  • 0
    @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. – 2017-02-16