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Question Number 128721 by rs4089 last updated on 09/Jan/21

∫_0 ^1 ((ln x)/(x(x^2 +1))) dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\:\mathrm{x}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:\mathrm{dx} \\ $$

Commented by Dwaipayan Shikari last updated on 09/Jan/21

(1/2)∫_0 ^1 ((logx)/x)((1/(x−i))+(1/(x+i)))dx  =(1/(2i))∫_0 ^1 ((logx)/x^2 )(((xi)/(x−i))+((xi)/(x+i)))dx  =(1/(2i))∫_0 ^1 ((logx)/x^2 )Σ_(n=1) ^∞ (xe^((π/2)i) )^n −(xe^(−(π/2)i) )^n   =∫_0 ^1 ((logx)/x^2 ).Σ_(n=1) ^∞ x^n ((sin((π/2)n))/n^2 )=Σ_(n≥1) ^∞ ((sin((π/2)n))/n^2 )∫_0 ^1 x^(n−2) log(x)dx  =((1/1^2 )−(1/3^2 )+(1/5^2 )−(1/7^2 )+...)∫_0 ^∞ e^(−(n−1)t) tdt=G(Σ_(n=1) ^∞ (1/((n−1)^2 )))→∞

$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{logx}}{{x}}\left(\frac{\mathrm{1}}{{x}−{i}}+\frac{\mathrm{1}}{{x}+{i}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{logx}}{{x}^{\mathrm{2}} }\left(\frac{{xi}}{{x}−{i}}+\frac{{xi}}{{x}+{i}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{logx}}{{x}^{\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({xe}^{\frac{\pi}{\mathrm{2}}{i}} \right)^{{n}} −\left({xe}^{−\frac{\pi}{\mathrm{2}}{i}} \right)^{{n}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{logx}}{{x}^{\mathrm{2}} }.\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}} \frac{{sin}\left(\frac{\pi}{\mathrm{2}}{n}\right)}{{n}^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\frac{\pi}{\mathrm{2}}{n}\right)}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{2}} {log}\left({x}\right){dx} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }+...\right)\int_{\mathrm{0}} ^{\infty} {e}^{−\left({n}−\mathrm{1}\right){t}} {tdt}={G}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }\right)\rightarrow\infty \\ $$

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