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Question Number 82517 by naka3546 last updated on 22/Feb/20

$$\underset{1}{\int }\stackrel{\mathrm{\infty }}{}\frac{\ln x}{x^2+1}dx,ConvergentorDivergent?$$

Commented by Tony Lin last updated on 22/Feb/20

$${\int }_1^{\mathrm{\infty }}\frac{lnx}{x^2+1}dx<{\int }_1^{\mathrm{\infty }}\frac{lnx}{x^2}dx$$$$\int \frac{lnx}{x^2}dx$$$$=-\frac{lnx}{x}-\int -\frac{1}{x^2}dx$$$$=\frac{-lnx-1}{x}+C$$$${\int }_1^{\mathrm{\infty }}\frac{lnx}{x^2}dx$$$$=-\underset{t\to \mathrm{\infty }}{\lim }\frac{lnt+1}{t}+1$$$$=-\underset{t\to \mathrm{\infty }}{\lim }\frac{1}{t}+1$$$$=1$$$$ByComparisonTest$$$${\int }_1^{\mathrm{\infty }}\frac{lnx}{x^2+1}dxisconvergent$$

Commented by mathmax by abdo last updated on 23/Feb/20

$$letA={\int }_1^{\mathrm{\infty }}\frac{lnx}{1+x^2}dxchangementx=\frac{1}{t}give$$$$A=-{\int }_0^1\frac{-ln\left(t\right)}{1+\frac{1}{t^2}}\times (-\frac{dt}{t^2})=-{\int }_0^1\frac{lnt}{1+t^2}dt$$$$=-{\int }_0^{1}ln\left(t\right)\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^{2n}\right)dt$$$$=-\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^1{t}^{2n}ln\left(t\right)dtbyparts$$$$u_n={\int }_0^1{t}^{2n}ln\left(t\right)dt={\left[\frac{1}{2n+1}{t}^{2n+1}ln\left(t\right)\right]}_0^1-{\int }_0^1\frac{1}{2n+1}{t}^{2n}dt$$$$=-\frac{1}{2n+1}{\left[\frac{1}{2n+1}{t}^{2n+1}\right]}_0^1=-\frac{1}{{(2n+1)}^2}\Rightarrow$$$$A=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(2n+1)}^2}andthisserieisconvergente.$$

Commented by mathmax by abdo last updated on 24/Feb/20

$$letI={\int }_1^{\mathrm{\infty }}\frac{ln\left(x\right)}{x^2+1}dxchangementx=tan\theta give$$$$I={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{ln\left(tan\theta \right)}{1+{tan}^2\theta }(1+{tan}^2\theta )d\theta ={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\{ln\left(sin\theta \right)-ln\left(cos\theta \right)\}d\theta$$$$={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}ln\left(sin\theta \right)d\theta -{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}ln\left(cos\theta \right)d\theta =H-K$$$$H+K={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}ln(\frac{1}{2}sin\left(2\theta \right)d\theta =-ln\left(2\right)\times \frac{\pi }{4}+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}ln\left(sin\left(2\theta \right)\right)d\theta$$$$=-\frac{\pi }{4}ln\left(2\right)+\frac{1}{2}{\int }_{\frac{\pi }{2}}^{\pi }ln\left(sinu\right)du(2\theta =u)$$$${\int }_{\frac{\pi }{2}}^{\pi }ln\left(sinu\right)du{=}_{u=\frac{\pi }{2}+\alpha }{\int }_0^{\frac{\pi }{2}}ln\left(cos\alpha \right)d\alpha =-\frac{\pi }{2}ln\left(2\right)\Rightarrow$$$$H+K=-\frac{\pi }{2}ln\left(2\right)$$$$K={\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}ln\left(cos\theta \right)d\theta {=}_{\theta =\frac{\pi }{2}-\alpha }{\int }_{\frac{\pi }{4}}^{0}ln\left(sin\alpha \right)(-d\alpha )$$$$={\int }_0^{\frac{\pi }{4}}ln\left(sin\alpha \right)d\alpha ....becontinued...$$

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