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Oct 19, 2012 at 14:29 comment added Noam D. Elkies Well anywhere you see products or sums of arbitrarily many variables (let alone multiple integrals of arbitrary multiplicity) there's an implicit use of induction to define and manipulate them...
Oct 18, 2012 at 21:19 history edited Daniel Miller CC BY-SA 3.0
Cleaned up the notation
Jul 6, 2010 at 22:22 comment added Eben Freeman I don't think you really need induction... after all, you have $\int_{[0,1]^n} \left( 1 - x_1 \cdots x_n \right)^{-1}dx_k = \int \sum_{i=0}^{\infty} (x_1 \cdots x_n)^i dx_k = \ldots = \sum_{i=1}^{\infty} 1/i^n$
Jul 6, 2010 at 19:58 history answered Daniel Miller CC BY-SA 2.5