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For n we define a double integral equation* Iₙ=124₀¹₀¹ -³ (xy) 1+xy (xy (1-xy) ) ⁿ \ dxdyequation* We denote dₙ=lcm (1, 2,. . . , n) and prove that for all n, equation* dₙ Iₙ= 15 (-1) ⁿ 2^n-4 dₙ (5) + (-1) ^n+1 dₙ ₑ=₀^n nr (₊=₁^n+r (-1) ^k-1k⁵) equation* Now if (5) is rational, then (5) =a/b, (a, b) =1 and a, b. Then we take n b so that b|dₙ. We show that for all n 1, 0<dₙ Iₙ<1. We denote equation* Sₙ=dₙ ₑ=₀^n nr (₊=₁^n+r (-1) ^k-1k⁵) equation* We prove for all n b, equation* dₙ Iₙ+ (-1) ^n \Sₙ\* where \x\=x-x denotes the fractional part of x and x denotes the greatest integer less than or equal to x. Later we show that the only integer value admissible is equation* dₙ Iₙ+ (-1) ^n \Sₙ\=0\ \ \ or\ \ \ dₙ Iₙ+ (-1) ^n \Sₙ\=1equation* Finally we prove that these above values are not possible. This gives a contradiction to our assumption that (5) is rational. Similarly for all m 3 and n 1, by defining equation* I₍, ₌=1 (2m+1) ₀¹₀¹ -^2m-1 (xy) 1+xy (xy (1-xy) ) ⁿ \ dxdyequation* where (s) denotes the Gamma function, we give an extension of the above proof to prove the irrationality of higher odd zeta values (2m+1) for all m 3
Shekhar Suman (Mon,) studied this question.