Question Number 120775 by mnjuly1970 last updated on 02/Nov/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Phi\overset{???} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({x}\right){tan}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:…{m}.{n}.\mathrm{1970}… \\ $$
Answered by Dwaipayan Shikari last updated on 02/Nov/20
![∫tan^(−1) x dx =xtan^(−1) x−∫(x/(1+x^2 )) =xtan^(−1) x−log((√(1+x^2 ))) (logx∫_0 ^1 tan^(−1) (x)dx=0) Φ=[log(x)∫_0 ^1 tan^(−1) x]−∫_0 ^1 tan^(−1) x+(1/2)∫_0 ^1 ((log(1+x^2 ))/x) Φ=−[xtan^(−1) x−log((√(1+x^2 )))]+(1/2)∫_0 ^1 Σ_(n=1) ^∞ (−1)^(n+1) x^(2n−1) .(1/n) Φ=−(π/4)+(1/2)log(2)+(1/2)Σ_(n=1) ^∞ (−1)^(n+1) ∫_0 ^1 x^(2n−1) .(1/n)dx Φ=−(π/4)+(1/2)log(2)+(1/2)Σ_(n=1) ^∞ (−1)^(n+1) (1/(2n^2 )) Φ=−(π/4)+(1/2)log(2)+(π^2 /(48))](https://www.tinkutara.com/question/Q120786.png)
$$\int{tan}^{−\mathrm{1}} {x}\:{dx} \\ $$$$={xtan}^{−\mathrm{1}} {x}−\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$={xtan}^{−\mathrm{1}} {x}−{log}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:\:\:\:\:\:\left({logx}\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left({x}\right){dx}=\mathrm{0}\right) \\ $$$$\Phi=\left[{log}\left({x}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {x}\right]−\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {x}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}} \\ $$$$\Phi=−\left[{xtan}^{−\mathrm{1}} {x}−{log}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right]+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{1}} .\frac{\mathrm{1}}{{n}} \\ $$$$\Phi=−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{1}} .\frac{\mathrm{1}}{{n}}{dx} \\ $$$$\Phi=−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$\Phi=−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$
Commented by mnjuly1970 last updated on 02/Nov/20
$${thank}\:{you}\:{so}\:{much}\:{mr}\:{dwaipayanexcellent} \\ $$
Commented by Dwaipayan Shikari last updated on 03/Nov/20
$${With}\:{pleasure} \\ $$
Answered by mindispower last updated on 02/Nov/20
![=[(xln(x)−x)tan^− (x)]_0 ^1 −∫_0 ^1 ((xln(x)−x)/(1+x^2 ))dx =−tan^− (1)+∫_0 ^1 (x/(1+x^2 ))dx−∫_0 ^1 (x/(1+x^2 ))ln(x)dx −(π/4)+((ln(2))/2)−Σ_(k≥0) ∫_0 ^1 (−1)^k x^(2k+1) ln(x)dx =−(π/4)+((ln(2))/2)−Σ_(k≥0) (−1)^k (−(1/((2k+2)^2 ))) =−(π/4)+((ln(2))/2)+(1/4)Σ_(k≥0) (((−1)^k )/((k+1)^2 )) =−(π/4)+((ln(2))/2)+(1/4)[(π^2 /(12))] =((−12π+π^2 +24ln(2))/(48))](https://www.tinkutara.com/question/Q120793.png)
$$=\left[\left({xln}\left({x}\right)−{x}\right){tan}^{−} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left({x}\right)−{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=−{tan}^{−} \left(\mathrm{1}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{ln}\left({x}\right){dx} \\ $$$$−\frac{\pi}{\mathrm{4}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}−\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}+\mathrm{1}} {ln}\left({x}\right){dx} \\ $$$$=−\frac{\pi}{\mathrm{4}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}−\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \left(−\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$$=−\frac{\pi}{\mathrm{4}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\pi}{\mathrm{4}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right] \\ $$$$=\frac{−\mathrm{12}\pi+\pi^{\mathrm{2}} +\mathrm{24}{ln}\left(\mathrm{2}\right)}{\mathrm{48}} \\ $$
Commented by mnjuly1970 last updated on 02/Nov/20
$${very}\:{nice}\:{sir}\:{mindspower} \\ $$$${grateful}.. \\ $$
Commented by mindispower last updated on 03/Nov/20
$${thanx}\:{withe}\:{pleasur}\:{nice}\:{day}\:{sir} \\ $$