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38
votes
Integral whose upper limit is the integral itself: $\int_{0}^{\int_{0}^{\ldots}\frac{1}{\sqrt{x}} \ \mathrm{d}x} \frac{1}{\sqrt{x}} \ \mathrm{d}x$
calculus
integration
definite-integrals
asked Jan 11, 2016 at 20:30
math.stackexchange.com
5
votes
Tensor product of rings: well-definedness of multiplication
abstract-algebra
ring-theory
tensor-products
algebras
asked Sep 2, 2021 at 16:19
math.stackexchange.com
Top Answers
No answers with score of 5 or more