# An integral in Gradshteyn and Ryzhik

Section 3.248 of the 4th edition of the table of integrals by Gradshteyn and Ryzhik contains three entries. They are of elementary examples of the beta function. In the 5th edition there are two new entries in this section: 3.248.5 is the integral from 0 to infinity of the the function $$\int_0^\infty \frac{1}{(1+x^2)^{3/2}\sqrt{f(x) + \sqrt{f(x)}}},$$
where
$$f(x) = 1 + \frac{4x^2}{3(1+x^2)^2.}$$ The answer is given as $\pi/(2 \sqrt{6})$. This is incorrect. The entry was taken out in the 7th edition. Now we know what the correct answer should be (a complicated difference of two elliptic integrals) and we also know if the inner square root is replaced by the 3/2 power, the answer is $\pi/(2 \sqrt{6})$. We would like to know if anyone has information about the origin of this entry.

The integral appeared in an edition of the table GR. I became interested in these evaluations and when we contacted the editors, there was no information about the origin of this formula. Since I had done some evaluations involving the double square root function, this entry caught my eye. I was hopeful that by bringing this entry to a wider forum, someone would know about its origin.

Thanks.

• Welcome, Victor! – Igor Rivin Sep 14 at 18:31
• Note: Victor Moll is one of the editors for recent editions of the book. – Gerald Edgar Sep 14 at 19:22
• But we want an exact answer. There is no point of Mathematica/Maple to evaluate integrals numerically. – Victor Moll Sep 15 at 1:42