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The function $e^{\frac{-1}{x(1-x)}}$ is a smooth positive function supported on the compact set $[0,1]$.

What is its integral? (According to WolframAlpha, it's approximately 0.007, but I was hoping for a more 'theoretical' understanding.)

Ryan
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    Here are some results for other bump functions that may be relevant: https://math.stackexchange.com/questions/145015/evaluate-definite-integral-int-11-exp1-x2-1-dx https://math.stackexchange.com/questions/585768/how-to-integrate-the-bump-functions-i-e-int-abe-frac1x-a-frac1x – whpowell96 Apr 22 '24 at 22:53

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The solution was given by @Harry Peter here.

Make $a=0$ and $b=1$ and then $$\int_0^1 e^{\frac{-1}{x(1-x)}}\,dx=\frac{2 }{e^2}\,\big(K_1(2)-K_0(2)\Big)$$ which is $0.00702986$ as given by Wolfram Alpha.

Claude Leibovici
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