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Question Number 165218 by Lordose last updated on 27/Jan/22

$$\mathrm{Question}\mathrm{by}\mathbf{M}.\mathbf{N}\mathbf{July}$$$$\mathrm{\Phi }={\int }_0^1\frac{\ln (1+{\mathrm{x}}^4+{\mathrm{x}}^8)}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Phi }\stackrel{\mathrm{x}={\mathrm{x}}^{\frac{1}{2}}}{=}\frac{1}{2}{\int }_0^1\frac{\ln (1+{\mathrm{x}}^2+{\mathrm{x}}^4)}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Phi }=\frac{1}{2}{\int }_0^1\frac{\ln \left(\frac{1-{\mathrm{x}}^2}{1-{\mathrm{x}}^6}\right)}{\mathrm{x}}\mathrm{dx}=\frac{1}{2}{\int }_0^1\frac{\ln (1-{\mathrm{x}}^2)}{\mathrm{x}}\mathrm{dx}-\frac{1}{2}{\int }_0^1\frac{\ln (1-{\mathrm{x}}^6)}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Phi }=\frac{1}{2}(\mathrm{A}-\mathrm{B})$$$$\mathrm{A}\stackrel{\mathrm{x}={\mathrm{x}}^{\frac{1}{2}}}{=}\frac{1}{2}{\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}=\frac{1}{2}{\mathbf{Li}}_2\left(1\right)$$$$\mathrm{B}\stackrel{\mathrm{x}={\mathrm{x}}^{\frac{1}{6}}}{=}\frac{1}{6}{\int }_0^1\frac{{\mathrm{x}}^{\frac{1}{6}-1}\ln (1-\mathrm{x})}{{\mathrm{x}}^{\frac{1}{6}}}\mathrm{dx}$$$$\mathrm{B}=\frac{1}{6}{\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}=\frac{1}{6}{\mathbf{Li}}_2\left(1\right)$$$$\mathrm{\Phi }=\frac{1}{2}(\frac{1}{2}{\mathbf{Li}}_2\left(1\right)-\frac{1}{6}{\mathbf{Li}}_2\left(1\right))=\frac{1}{6}{\mathbf{Li}}_2\left(1\right)$$$$\mathbf{\Phi }=\frac{\mathit{\zeta }\left(2\right)}{3}▴▴▴$$

Commented by mnjuly1970 last updated on 27/Jan/22

$$thxalotsirlordose$$

Answered by Ar Brandon last updated on 27/Jan/22

$$\mathrm{\Phi }={\int }_0^1\frac{\ln (1+x^4+x^8)}{x}dx={\int }_0^1\frac{1}{x}\ln \left(\frac{1-x^{12}}{1-x^4}\right)dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^1(x^{4n-1}-x^{12n-1})dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}(\frac{1}{4n}-\frac{1}{12n})$$$$=\frac{1}{6}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{\zeta \left(2\right)}{6}=\frac{{\pi }^2}{36}$$

Commented by amin96 last updated on 28/Jan/22

$$\mathbf{correct}\mathbf{sir}.\mathbf{great}\mathbf{solution}$$

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