How to prove that $$\int_{0}^{1}\log^2\Gamma(q) dq=\frac{\gamma^2}{12}+\frac{\pi^2}{48}+\frac{\gamma}{3}L_1+\frac{4}{3}L_1^2-(\gamma+2L_1)\frac{\zeta''(2)}{\pi^2}+\frac{\zeta'(2)}{2\pi^2}$$


$$L_1=\int_{0}^{1}\log\Gamma(q) dq=\log\sqrt{2\pi}$$




closed as off-topic by Carl Mummert, Claude Leibovici, mrp, Parcly Taxel, R_D Oct 4 '16 at 9:25

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  • 1
    $\begingroup$ Does $\log^2 \Gamma(q)$ mean $\log \bigl(\log \Gamma(q)\bigr)$ or $\bigl(\log \Gamma(q)\bigr)^2$? $\endgroup$ – Daniel Fischer Oct 3 '16 at 12:05
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    $\begingroup$ And where does this come from, please? $\endgroup$ – peter a g Oct 3 '16 at 12:06
  • $\begingroup$ @DanielFischer $\log^2\Gamma(q)$ mean $\log(\Gamma(q))^2$ $\endgroup$ – vito Oct 3 '16 at 12:08
  • $\begingroup$ @peterag see here, page 5 $\endgroup$ – vito Oct 3 '16 at 12:10
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    $\begingroup$ Use this $$\log{\Gamma(x)}=\frac{1}{2}\log(2\pi)+\sum^\infty_{n=1}\left\{\frac{1}{2n}\cos(2\pi nx)+\frac{\gamma+\log(2\pi n)}{n\pi}\sin(2\pi nx)\right\}$$ and the orthogonality of trigonometric functions to conclude. $\endgroup$ – M.N.C.E. Oct 4 '16 at 1:33