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Question Number 110543 by mnjuly1970 last updated on 29/Aug/20

Answered by mathdave last updated on 29/Aug/20

$$solution$$$$lety=x^{2}anddx=\frac{1}{2\sqrt{y}}dythenputtinginto$$$$I=\frac{1}{2}{\int }_0^1\frac{\ln (1-y)\ln (1+y)}{y}dy$$$$fromalgebraicudingthisnotation$$$$ab=\frac{1}{4}{(a+b)}^2-\frac{1}{4}{(a-b)}^2$$$$\ln (1-y)\ln (1+y)=\frac{1}{4}{\ln }^2(1-y^2)-\frac{1}{4}{\ln }^2\left(\frac{1-y}{1+y}\right)$$$$I=\frac{1}{8}{\int }_0^1\frac{{\ln }^2(1-y^2)}{y}dy-\frac{1}{8}{\int }_0^1\frac{{\ln }^2\left(\frac{1-y}{1+y}\right)}{y}dy$$$$let{I}_x={\int }_0^1\frac{{\ln }^2(1-y^2)}{y}dy(letx=1-y^2$$$$anddy=-\frac{dx}{2\sqrt{1-x}}$$$$I_x=\frac{1}{2}{\int }_0^1\frac{{\ln }^{2}x}{1-x}dx$$$$let$$$$I_x\left(a\right)=\frac{1}{2}{\int }_0^1\frac{{\ln }^{2}x}{1-x}dx=\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^{n-1}{\ln }^{2}xdx$$$$\frac{{\partial }^2}{\partial a^2}{\mid }_{a=0}{I}_x\left(a\right)=\frac{1}{2}\frac{{\partial }^2}{\partial a^2}\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^{n-1+a}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}$$$$I_x^{{}^{″}}\left(0\right)=\zeta \left(3\right)......\left(x\right)$$$$let{I}_{xx}={\int }_0^1\frac{{\ln }^2\left(\frac{1-y}{1+y}\right)}{y}dyletx=\frac{1-y}{1+y}anddy=-\frac{2}{{(1+x)}^2}$$$$I_{xx}=2{\int }_0^1\frac{{\ln }^{2}x}{1-x^2}dx=2{\int }_0^1\frac{{\ln }^2\left(x\right)}{(1+x)(1-x)}dx$$$$I_{xx}={\int }_0^1\frac{{\ln }^{2}x}{1+x}dx+{\int }_0^1\frac{{\ln }^{2}x}{1-x}dxtherefore$$$$I_{xx}^{{}^{″}}\left(0\right)={(-1)}^{n}2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}+2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}$$$$I_{xx}^{{}^{″}}\left(0\right)=2\times \frac{3}{4}\zeta \left(3\right)+2\zeta \left(3\right)=\frac{7}{2}\zeta \left(3\right).......\left(2\right)$$$$but$$$$I=\frac{1}{8}[I_x^{{}^{″}}\left(0\right)-I_{xx}^{{}^{″}}\left(0\right)]=\frac{1}{8}[\zeta \left(3\right)-\frac{7}{2}\zeta \left(3\right)]$$$$I=\frac{1}{8}[-\frac{5}{2}\zeta \left(3\right)]=-\frac{5}{16}\zeta \left(3\right)$$$$\because {\int }_0^1\frac{\ln (1-x^2)\ln (1+x^2)}{x}dx=-\frac{5}{16}\zeta \left(3\right)$$$$where\zeta \left(3\right)\Rightarrow {apery}^{\prime }sconstant$$$$bymathdave\left(29/08/2020\right)$$

Commented by mnjuly1970 last updated on 29/Aug/20

$$thankyouverymuch...sir...$$

Commented by mathdave last updated on 29/Aug/20

$$youarewelcome$$

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