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Question Number 136626 by mnjuly1970 last updated on 24/Mar/21

$$..........nicecalculus.........$$$$supposethat:::$$$$\phi \left(p\right)={\int }_0^{\mathrm{\infty }}\frac{ln(1+x)}{{(p+x)}^2}...✓$$$$findthevalueof::$$$${\underset{0}{\int }}^1\frac{\phi \left(p\right)}{1+p}dp=?...$$$$$$

Answered by mindispower last updated on 24/Mar/21

$$\phi \left(p\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{(p+x)(1+x)}=\frac{1}{p-1}{\left[ln\left(\frac{1+x}{p+x}\right)\right]}_0^{\mathrm{\infty }}$$$$=\frac{ln\left(p\right)}{p-1}$$$${\int }_0^1\frac{-ln\left(p\right)}{(1-p^2)}dp=-{\int }_0^1\sum _{n⩾0}{p}^{2n}ln\left(p\right)dp$$$$=\mathrm{\Sigma }\frac{1}{{(2n+1)}^2}=\frac{3}{4}\mathit{\zeta }\left(2\right)=\frac{{\pi }^2}{8}$$$$$$

Commented by mnjuly1970 last updated on 24/Mar/21

$$tashakor.grateful..sirpower..$$

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