Is it possible to compute the convolution of $\mathbb{1}_{\mathbb{Q}}$ with $e^{-1/(1-t)^2}$?
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Since $1_\mathbb{Q}(x) = 0$ for ae. $x$ [$m$], the answer is yes, and the value is zero. – copper.hat Feb 13 '14 at 04:13
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@copper.hat do the singularities of the other function in the convolution make a difference? Certainly, $\mathbb{1_Q}*\delta \neq 0$ – Ben Grossmann Feb 13 '14 at 04:16
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@Omnomnomnom: The other function is very well behaved. (It can be taken to be 0 for $t=1$ which makes it continuous, but this in unnecessary. – copper.hat Feb 13 '14 at 04:19
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@copper.hat ah, indeed, I forgot about the minus sign in the numerator. – Ben Grossmann Feb 13 '14 at 04:21
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1@Omnomnomnom: However, if it the minus sign was absent, the product is still zero ae. so the integral is still defined and zero. – copper.hat Feb 13 '14 at 04:23