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Please help me to find a closed form for this integral: $$\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx$$

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  • $\begingroup$ Did you obtain this sum from an integral? If yes, what was its form? $\endgroup$ – Start wearing purple May 8 '13 at 20:42
  • $\begingroup$ Yes, this sum is part of a large expression that came up after a long chain of transformations of the integral $\int_0^\infty\frac{\arctan(x)\log(1+x^2)\sqrt{x}}{1+x^2}$ $\endgroup$ – Laila Podlesny May 8 '13 at 20:50
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    $\begingroup$ I moved the question about the sum $\sum_{n=1}^\infty\frac{\psi(n+\frac{5}{4})}{(1+2n)(1+4n)^2}$ to math.stackexchange.com/questions/386142/… to make it possible to accept answers both about the integral and the sum. $\endgroup$ – Laila Podlesny May 9 '13 at 1:19
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The integral can be trasformed into a "computable" form. Indeed, \begin{align} I&=\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx=\\ &=2\int_0^{\infty}\frac{\arctan y^2\ln(1+y^4)}{1+y^4}y^2dy=\\ &=\frac{1}{2i}\int_{-\infty}^{\infty}\frac{\ln^2(1+iy^2)-\ln^2(1-iy^2)}{1+y^4}y^2dy=\\ &=\frac{1}{4}\int_{-\infty}^{\infty}\left(\frac{1}{1+iy^2}-\frac{1}{1-iy^2}\right)\Bigl(\ln^2(1+iy^2)-\ln^2(1-iy^2)\Bigr)dy=\\ &=\frac12\mathrm{Re}\int_{-\infty}^{\infty}\frac{\ln^2(1+iy^2)-\ln^2(1-iy^2)}{1+iy^2}dy, \end{align} where the logarithms are defined on their main sheets. The remaining integrals can be evaluated by suitably deforming the contours (or using Mathematica). The final result is $$I=\frac{\pi}{6\sqrt{2}}\left(12G+9\ln^22+3\pi\ln2+\pi^2\right)\approx 11.7433,$$ where $G$ denotes the Catalan's constant.

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