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arXiv 2018-01-29 2 views

True value of an integral in Gradshteyn and Ryzhik's table

de Reyna, J. Arias

Original · EN

Victor Moll pointed out that entry 3.248.5 in the sixth edition of Gradshteyn and Ryzhik tables of integrals was incorrect. He asked some years ago what was the true value of this integral. I evaluate it in terms of two elliptic integrals. The evaluation is standard but involved, using real and complex analysis. I prove ∫₀∞dx(1+x²)³/²[φ(x)+√φ(x)]¹/²= √3-1√2Π(π/2,k,3⁻¹/²)-1√2F(α, 3⁻¹/²), where φ(x)=1+4x²/2(1+x²)², k=2-√3, α=√k and F(φ,k), Π(φ,n²,k) the elliptic integrals of the first and third kind respectively.

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