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Question Number 130254 by Eric002 last updated on 23/Jan/21

$$prove$$$${\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(x\right)ln(1+x^2)}{1+x^2}dx=\frac{7\pi }{4}\zeta \left(3\right)+\frac{{\pi }^{3}ln\left(2\right)}{4}$$$$where\zeta \left(3\right)isa{pery}^{,}sconstant$$

Answered by mindispower last updated on 26/Jan/21

$${\int }_0^{\frac{\pi }{2}}({ln}^2\left(sin\left(x\right)\right)+{ln}^2\left({cos}^2\left(x\right)\right)-2ln\left(sin\left(x\right)\right)ln\left(cos\left(x\right)\right).ln\left(\frac{1}{{cos}^2\left(x\right)}\right)dx$$$$=-2{\int }_0^{\frac{\pi }{2}}{ln}^2\left(sin\left(x\right)\right)ln\left(cos\left(x\right)\right)dx-2{\int }_0^{\frac{\pi }{2}}{ln}^3\left(cos\left(x\right)\right)dx$$$$+4{\int }_0^{\frac{\pi }{2}}{ln}^2\left(cos\left(x\right)\right)ln\left(sin\left(x\right)\right)dx=-2A-2B+4C$$$$recall$$$$\beta (a,b)=2{\int }_0^{\frac{\pi }{2}}{cos}^{2a-1}\left(x\right){sin}^{2b-1}\left(x\right)dx$$$$\frac{{\partial }^3}{{\partial }^{2}b\partial a}\beta (\frac{1}{2},\frac{1}{2})=8A,\frac{{\partial }^3}{{\partial }^{2}a\partial b}\beta (\frac{1}{2},\frac{1}{2})=8C$$$$\frac{{\partial }^3\beta (\frac{1}{2},\frac{1}{2})}{{\partial }^{3}a}=8B$$$${\partial }_a\beta (a,b)=\beta (a,b)(\mathrm{\Psi }\left(a\right)-\mathrm{\Psi }(a+b))$$$${\partial }_a^2\beta (a,b)=\beta (a,b)(\mathrm{\Psi }\left(a\right)-\mathrm{\Psi }{(a+b)}^2+({\mathrm{\Psi }}^{\prime }\left(a\right)-{\mathrm{\Psi }}^{\prime }(a+b))\beta (a,b)$$$${\partial }_a^2{\partial }_b\beta (a,b)=\beta (a,b)(\mathrm{\Psi }\left(b\right)-\mathrm{\Psi }(a+b)){(\mathrm{\Psi }\left(a\right)-\mathrm{\Psi }(a+b))}^2$$$$-2{\mathrm{\Psi }}^{\prime }(a+b)(\mathrm{\Psi }\left(a\right)-\mathrm{\Psi }(a+b))(\mathrm{\Psi }\left(a\right)-\mathrm{\Psi }(a+b))\beta (a,b)$$$$+\beta (a,b)(\mathrm{\Psi }\left(b\right)-\mathrm{\Psi }(a+b)))({\mathrm{\Psi }}^{\prime }\left(a\right)-{\mathrm{\Psi }}^{\prime }(a+b))$$$$-{\mathrm{\Psi }}^{″}(a+b)\beta (a,b))$$$${\partial }^{2}a\partial b\beta (\frac{1}{2_{}},\frac{1}{2})$$$$iwillfinish?itlater$$$$\mathrm{\Psi }\left(\frac{1}{2}\right)caneaslyvaluted$$$${\mathrm{\Psi }}^{\prime }\left(1\right)=\zeta \left(2\right)=\frac{{\pi }^2}{6}...=$$

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