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Question Number 114403 by bemath last updated on 19/Sep/20

$$\int \frac{\ln (1+x^4)}{x}dx$$$$$$

Commented by Dwaipayan Shikari last updated on 19/Sep/20

$${\int }_0^1{(-1)}^n.\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{4n-1}}{n}dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^1\frac{x^{4n-1}}{n}dx$$$$\sum ^{\mathrm{\infty }}{(-1)}^n{\left[\frac{x^{4n}}{4n.n}\right]}_0^1$$$$\frac{1}{4}\sum ^{\mathrm{\infty }}{(-1)}^n\frac{1}{n^2}=\frac{{\pi }^2}{48}$$

Answered by mathmax by abdo last updated on 19/Sep/20

$$\mathrm{if}\mathrm{the}\mathrm{Q}\mathrm{is}{\int }_0^1\frac{\ln (1+{\mathrm{x}}^4)}{\mathrm{x}}\mathrm{dx}\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{du}}\ln (1+\mathrm{u})=\frac{1}{1+\mathrm{u}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}$$$$\Rightarrow \ln (1+\mathrm{u})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}$$$$\Rightarrow \ln (1+{\mathrm{x}}^4)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{x}}^{4\mathrm{n}}}{\mathrm{n}}\Rightarrow \frac{\ln (1+{\mathrm{x}}^4)}{\mathrm{x}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\mathrm{x}}^{4\mathrm{n}-1}$$$${\int }_0^1\frac{\ln (1+{\mathrm{x}}^4)}{\mathrm{x}}\mathrm{dx}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{4\mathrm{n}-1}\mathrm{dx}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}\times \frac{1}{4\mathrm{n}}=-\frac{1}{4}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}$$$$=-\frac{1}{4}(2^{1-2}-1)\xi \left(2\right)=-\frac{1}{4}\times (-\frac{1}{2})\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{48}$$$${\int }_0^1\frac{\ln (1+{\mathrm{x}}^4)}{\mathrm{x}}\mathrm{dx}=\frac{{\pi }^2}{48}$$

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