Question Number 211857 by Frix last updated on 22/Sep/24
$$\underset{\mathrm{1}} {\overset{\infty} {\int}}\left(\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}}×\frac{\mathrm{ln}\:{x}}{{x}}\right){dx}=? \\ $$
Answered by Berbere last updated on 23/Sep/24
![∫_1 ^∞ ((tan^(−1) (x)ln(x))/x^2 )dx;u′=(1/x^2 );v=ln(x)tan^(−1) (x) =[−((ln(x)tan^(−1) (x))/x)]_1 ^∞ +∫_1 ^∞ [((tan^(−1) (x))/x^2 )+((ln(x))/(x(1+x^2 )))]_1 ^∞ dx =∫_1 ^∞ ((tan^(−1) (x))/x^2 )dx+∫_1 ^∞ ((ln(x))/(x(1+x^2 )))dx=A+B A=[−((tan^(−1) (x))/x)]_1 ^∞ +∫_1 ^∞ (1/(x(x^2 +1)))dx;x→(1/x) =(π/4)+∫_0 ^1 (x/(x^2 +1))=(π/4)+((ln(5))/2) B=−∫_0 ^1 x((ln(x))/(1+x^2 ))dx=−(1/2)∫_0 ^1 ((ln(x))/(1+x))dx...x→x^2 =(1/2)∫_0 ^1 ((ln(1+x))/x)dx=(1/2)∫_0 ^1 Σ_(n≥1) (((−1)^(n−1) )/n)x^(n−1) dx =(1/2)Σ(((−1)^(n−1) )/n^2 )=−(1/2)Li_2 (−1)=(π^2 /(24)) J=(π^2 /(24))+(π/4)+((ln(5))/2)](https://www.tinkutara.com/question/Q211876.png)
$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right){ln}\left({x}\right)}{{x}^{\mathrm{2}} }{dx};{u}'=\frac{\mathrm{1}}{{x}^{\mathrm{2}} };{v}={ln}\left({x}\right)\mathrm{tan}^{−\mathrm{1}} \left({x}\right) \\ $$$$=\left[−\frac{{ln}\left({x}\right)\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}\right]_{\mathrm{1}} ^{\infty} +\int_{\mathrm{1}} ^{\infty} \left[\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{2}} }+\frac{{ln}\left({x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\right]_{\mathrm{1}} ^{\infty} {dx} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx}+\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}={A}+{B} \\ $$$${A}=\left[−\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}\right]_{\mathrm{1}} ^{\infty} +\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx};{x}\rightarrow\frac{\mathrm{1}}{{x}} \\ $$$$=\frac{\pi}{\mathrm{4}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}=\frac{\pi}{\mathrm{4}}+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}} \\ $$$${B}=−\int_{\mathrm{0}} ^{\mathrm{1}} {x}\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}{dx}…{x}\rightarrow{x}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}−\mathrm{1}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\Sigma\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$${J}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\pi}{\mathrm{4}}+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Frix last updated on 23/Sep/24
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