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How to find the value of $$I_1=\int_0^\infty\frac{\sqrt{x}\arctan{x}\log^2({1+x^2})}{1+x^2}dx$$ If we put $$I_2=\int_0^\infty\frac{\arctan^2({x})\log({1+x^2})}{\sqrt{x}(1+x^2)}dx$$ After long calculations,I have found the following relationship. $$I_1-2I_2=\frac{\pi\sqrt{2}}{4}\left(\frac{5}{4}{\pi^3}-4{\pi}G+4{\pi}\ln^{2}2+\frac{5}{2}{\pi^2}\ln2+24G\ln2+12\ln^{3}2+\frac{27}{2}{\zeta(3)}\right)$$ Where $G$ is Catalan's number.

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Sketch of a particular method for integral 1.

Consider the series \begin{align} \tan^{-1}(x) = \sum_{n=0}^{\infty} \frac{(1)_{n} (1)_{n}}{n! \, (3/2)_{n}} \, \frac{x^{2n+1}}{(1+x^{2})^{n+1}}. \end{align} Now consider the integral \begin{align} I(\mu, \nu) = \int_{0}^{\infty} \frac{x^{\nu} \, tan^{-1}(x) }{ (1+x^{2})^{\mu +1}} \, dx. \end{align} From this it is seen that: \begin{align} I(\mu, \nu) &= \sum_{n=0}^{\infty} \frac{ (1)_{n} (1)_{n}}{n! (3/2)_{n}} \, \int_{0}^{\infty} \frac{x^{2n+\nu+1} \, dx}{(1+x^{2})^{n+\mu+2}} \\ &= \frac{1}{2} \, \sum_{n=0}^{\infty} \frac{ (1)_{n} (1)_{n}}{n! (3/2)_{n}} \, B(n+ \nu/2 +1, \mu -\nu/2 +1) \\ &= \frac{1}{2} \, B\left( \frac{\nu}{2} +1, \mu - \frac{\nu}{2} + 1\right) \, {}_{3}F_{1}\left( 1, 1, \frac{\nu}{2}+1; \frac{3}{2}, \mu+2; 1\right). \end{align} For the case of $\nu = 1/2$ this reduces to \begin{align} \int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) }{ (1+x^{2})^{\mu +1}} \, dx = \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\mu + \frac{3}{4}\right)}{2 \Gamma(\mu + 2)} {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{4}, \mu+2; 1\right) \end{align} Taking the derivative of both sides with respect to $\mu$ leads to \begin{align} \int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) \, \ln(1+x^{2})}{ (1+x^{2})^{\mu +1}} \, dx &= \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\mu + \frac{3}{2}\right)}{2 \, \Gamma(\mu + 2)} \, \left[ \psi\left( \mu + \frac{3}{2}\right) - \psi(\mu+2) \right] \, {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2}, \mu + 2; 1 \right) \\ & \hspace{10mm} + \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left( \mu + \frac{3}{2}\right)}{2 \, \Gamma(\mu + 2) } \, \partial_{\mu} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right] \end{align} and evaluating at $\mu =0$ leads to \begin{align} \int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) \, \ln(1+x^{2})}{ (1+x^{2})^{\mu +1}} \, dx &= \frac{\pi}{4\sqrt{2}} \left( 1 + 3\ln(2) - \frac{\pi}{2} \right) \, {}_{3}F_{2}\left( 1 ,1, \frac{5}{4}; \frac{3}{2}, 2; 1 \right) \\ & \hspace{10mm} - \frac{\pi}{4 \sqrt{2}} \partial_{\mu} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2}, \mu+2; 1\right) \right]_{\mu=0} \end{align}

Repeating this process leads to \begin{align} \int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) \, \ln^{2}(1+x^{2})}{ (1+x^{2})^{\mu +1}} \, dx &= \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\mu + \frac{3}{2}\right)}{2 \, \Gamma(\mu + 2)} \, \left[ \psi^{'}\left( \mu + \frac{3}{2}\right) - \psi^{'}(\mu+2) + \left( \psi\left(\mu + \frac{3}{2}\right) - \psi(\mu+2) \right)^{2} \right] \\ & \hspace{10mm} \cdot \, {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2}, \mu + 2; 1 \right) \\ & \hspace{2mm} + \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left( \mu + \frac{3}{2}\right)}{ \Gamma(\mu + 2) } \, \left[ \psi\left(\mu + \frac{3}{2}\right) - \psi(\mu+2) \right] \, \partial_{\mu} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right] \\ & \hspace{10mm} + \frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left( \mu + \frac{3}{2}\right)}{ \Gamma(\mu + 2) } \, \partial_{\mu}^{2} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right] \end{align} When $\mu = 0$ this reduces to \begin{align} \int_{0}^{\infty} \frac{\sqrt{x} \, tan^{-1}(x) \, \ln^{2}(1+x^{2})}{1+x^{2}} \, dx &= \frac{\pi}{4 \sqrt{2}} \, \left[ - 8 G + 2 - \pi + (6 - 3 \pi) \ln(2) + 9 \ln^{2}(2) + \frac{13 \pi^{2}}{12} \right] \\ & \hspace{10mm} \cdot \, {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2}, 2; 1 \right) \\ & + \frac{\pi}{ 2\sqrt{2} } \, \left[ \frac{\pi}{2} - 1 - 3 \ln(2) \right] \, \partial_{\mu} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right]_{\mu = 0} \\ & \hspace{10mm} + \frac{\pi}{ 4 \sqrt{2} } \, \partial_{\mu}^{2} \left[ {}_{3}F_{2}\left( 1, 1, \frac{5}{4}; \frac{3}{2} , \mu + 2; 1 \right) \right]_{\mu = 0} \end{align}

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  • $\begingroup$ Amazing! Small point: The logarithm is squared in the question, but not in your integral. This "just" means one should differentiate once more... $\endgroup$ – mickep Mar 21 '15 at 20:50
  • $\begingroup$ @mickep I have added the remaining work to be more along the lines of what the integral actually stated. $\endgroup$ – Leucippus Mar 21 '15 at 22:49
  • $\begingroup$ Nice use of the hypergeometric series for arctangent. I wonder whether the hypergeometrics can be differentiated explicitly to yield anything useful. (I did once see a paper about how do do that, but it was so general I've forgotten the procedure and the results...) $\endgroup$ – Chappers Mar 22 '15 at 1:51
  • $\begingroup$ @Leucippus.Very nice solution +1.I believed in a simple result? $\endgroup$ – user178256 Mar 22 '15 at 16:12

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