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Question Number 143109 by mnjuly1970 last updated on 10/Jun/21

Commented by mnjuly1970 last updated on 10/Jun/21

$$prove{\mathrm{I}}_1=-\frac{{\pi }^2}{192}.....?$$

Answered by qaz last updated on 10/Jun/21

$${\mathrm{I}}_1={\int }_0^1\frac{{\mathrm{x}}^3}{1+{\mathrm{x}}^4}\mathrm{lnxdx}$$$$=\frac{1}{4}[\ln (1+{\mathrm{x}}^4)\mathrm{lnx}{\mid }_0^1-{\int }_0^1\frac{\ln (1+{\mathrm{x}}^4)}{\mathrm{x}}\mathrm{dx}]$$$$=-\frac{1}{4}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{4\mathrm{n}-1}\mathrm{dx}$$$$=-\frac{1}{4}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}-1}}{{4\mathrm{n}}^2}$$$$=-\frac{1}{16}(1-2^{1-2})\zeta \left(2\right)$$$$=-\frac{{\pi }^2}{192}$$

Commented by mnjuly1970 last updated on 10/Jun/21

$$thanksalot...$$

Answered by Ar Brandon last updated on 10/Jun/21

$${\mathrm{I}}_1=\frac{1}{4}{[\ln (1+{\mathrm{x}}^4)\mathrm{lnx}-\int \frac{\ln (1+{\mathrm{x}}^4)}{\mathrm{x}}\mathrm{dx}]}_0^1$$$$=-\frac{1}{4}{\int }_0^1\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}+1}\frac{{\mathrm{x}}^{4\mathrm{n}-1}}{\mathrm{n}}=-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{{16\mathrm{n}}^2}$$$$=-\frac{\eta \left(2\right)}{16}=-\frac{\zeta \left(2\right)}{16}(1-\frac{1}{2})=-\frac{{\pi }^2}{192}$$

Commented by mnjuly1970 last updated on 10/Jun/21

$$grateful..$$

Answered by Olaf_Thorendsen last updated on 10/Jun/21

$${\mathrm{I}}_1={\int }_0^1\frac{x^3}{1+x^4}\ln xdx$$$${\mathrm{I}}_1={\int }_0^1{x}^3\ln x\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{x}^{4n}dx$$$${\mathrm{I}}_1={\int }_0^1\ln x\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{x}^{4n+3}dx$$$${\mathrm{I}}_1={\left[\ln x\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{x^{4n+4}}{4n+4}\right]}_0^1$$$$-{\int }_0^1\frac{1}{x}\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{x^{4n+4}}{4n+4}dx$$$${\mathrm{I}}_1=-{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{x^{4n+3}}{4n+4}dx$$$${\mathrm{I}}_1=-{\left[\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{x^{4n+4}}{{(4n+4)}^2}\right]}_0^1$$$${\mathrm{I}}_1=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(4n+4)}^2}$$$${\mathrm{I}}_1=\frac{1}{16}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{{(n+1)}^2}$$$${\mathrm{I}}_1=\frac{1}{16}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2}=\frac{1}{16}(-\frac{{\pi }^2}{12})=-\frac{{\pi }^2}{192}$$

Commented by mnjuly1970 last updated on 10/Jun/21

$$merceymrolaf...$$

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