Integrating over a hypercube, not a hypersphere
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Denote $square_m={pmb{x}=(x_1,dots,x_m)inmathbb{R}^m: 0leq x_ileq1,,,forall i}$ be an $m$ -dimensional cube. It is all too familiar that $int_{square_1}frac{dx}{1+x^2}=frac{pi}4$ . QUESTION. If $VertcdotVert$ stands for the Euclidean norm, then is this true? $$int_{square_{2n-1}}frac{dpmb{x}}{(1+Vertpmb{x}Vert^2)^n} =frac{pi^n}{4^nn!}.$$
mg.metric-geometry soft-question integration
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asked Dec 19 '18 at 17:06
T. Amdeberhan
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