Gradshteyn和Ryzhik表中积分的真值
True value of an integral in Gradshteyn and Ryzhik's table
摘要：
维克多莫尔指出，条目3.248.5在第六版的Gradshteyn和Ryzhik表的积分是不正确的。他问几年前什么是真正的价值，这一积分。我用两个椭圆积分来计算它。评价是标准的，但涉及，使用真实和复分析。证明了$$\int_0^{\infty}\frac{dx}{(1+x^2)^{3/2}[φ(x)+\sqrt{φ(x)}]^{1/2}}= \frac{\sqrt{3}-1}{\sqrt{2}}Π(π/2,k,3^{-1/2})-\frac{1}{\sqrt{2}}F(α, 3^{-1/2}),$$，其中$φ(x)=1+4x^2/2(1+x^2)^2$，$k=2-\sqrt{3}$，$α=\arcsin\sqrt{k}$和$F(φ,k)$，$Π(φ,n^2,k)$分别是第一类和第三类椭圆积分.
英文摘要：
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 $$\int_0^{\infty}\frac{dx}{(1+x^2)^{3/2}[φ(x)+\sqrt{φ(x)}]^{1/2}}= \frac{\sqrt{3}-1}{\sqrt{2}}Π(π/2,k,3^{-1/2})-\frac{1}{\sqrt{2}}F(α, 3^{-1/2}),$$ where $φ(x)=1+4x^2/2(1+x^2)^2$, $k=2-\sqrt{3}$, $α=\arcsin\sqrt{k}$ and $F(φ,k)$, $Π(φ,n^2,k)$ the elliptic integrals of the first and third kind respectively.
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