Question Number 138283 by mnjuly1970 last updated on 11/Apr/21
![......advanced ........... calculus...... prove that:: 𝛗=∫_0 ^( 1) ((ln(1+x^2 ).arctan(x))/x^2 )dx= proof::: 𝛗=_(⟨substitution⟩) ^(x=tan(θ)) ∫_0 ^( (π/4)) ((ln(1+tan^2 (θ)).θ)/(tan^2 (θ)))(1+tan^2 (θ))dθ =^(⟨simplification⟩) ∫_0 ^( (π/4)) ((θ.ln((1/(cos^2 (θ)))))/(sin^2 (θ)))dθ =−2∫_0 ^( (π/4)) ((θ.ln(cos(θ))/(sin^2 (θ)))dθ =^(i.b.p) 2{[(cot(θ).θ.ln(cos(θ))]_0 ^(π/4) −∫_0 ^( (π/4)) (cot(θ).[ln(cos(θ))−θ.tan(θ)]dθ =2.(π/4).ln(((√2)/2))−2∫_0 ^( (π/4)) cot(θ).ln(cos(θ))dθ+2∫_0 ^( (π/4)) θdθ =((−π)/4)ln(2)−Φ+(π^2 /(16)) Φ=∫_0 ^( (π/4)) ((cos(θ))/(sin(θ))).ln(1−sin^2 (θ))dθ =^(sin(θ)=y) ∫_0 ^( ((√2)/2)) ((ln(1−y^2 ))/y)dy=−∫_0 ^( ((√2)/2)) Σ_(n=1) ^∞ (y^(2n−1) /n)dy =−Σ[(y^(2n) /(2n^2 ))]_0 ^( ((√2)/2)) =((−1)/2) li_2 ((1/2)) =((−1)/2){(π^2 /(12))−(1/2)ln^2 (2)}=((−π^2 )/(24))+(1/4)ln^2 (2)... ∴ 𝛗=((−π)/4)ln(2)+(π^2 /(24))+(1/4)ln^2 (2)+(π^2 /(16)) ......... 𝛗 =((5π^2 )/(48))−((12π)/(48))ln(2)+((12)/(48)) ln^2 (2) ..... ........𝛗=(1/(48)){5π^2 −12πln(2)+12ln^2 (2)}](https://www.tinkutara.com/question/Q138283.png)
$$\:\:\:\:\:\:\:\:\:\:\:……{advanced}\:\:\:………..\:\:{calculus}…… \\ $$$$\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right).{arctan}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}= \\ $$$$\:{proof}::: \\ $$$$\:\:\:\boldsymbol{\phi}\underset{\langle{substitution}\rangle} {\overset{{x}={tan}\left(\theta\right)} {=}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\theta\right)\right).\theta}{{tan}^{\mathrm{2}} \left(\theta\right)}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\theta\right)\right){d}\theta \\ $$$$\overset{\langle{simplification}\rangle} {=}\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\theta.{ln}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\theta\right)}\right)}{{sin}^{\mathrm{2}} \left(\theta\right)}{d}\theta \\ $$$$\:\:\:\:\:\:\:=−\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\theta.{ln}\left({cos}\left(\theta\right)\right.}{{sin}^{\mathrm{2}} \left(\theta\right)}{d}\theta \\ $$$$\:\:\:\:\:\overset{{i}.{b}.{p}} {=}\mathrm{2}\left\{\left[\left({cot}\left(\theta\right).\theta.{ln}\left({cos}\left(\theta\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \left({cot}\left(\theta\right).\left[{ln}\left({cos}\left(\theta\right)\right)−\theta.{tan}\left(\theta\right)\right]{d}\theta\right.\right.\right. \\ $$$$\:\:\:\:\:\:=\mathrm{2}.\frac{\pi}{\mathrm{4}}.{ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)−\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {cot}\left(\theta\right).{ln}\left({cos}\left(\theta\right)\right){d}\theta+\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \theta{d}\theta \\ $$$$\:\:\:\:\:\:=\frac{−\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\Phi+\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{cos}\left(\theta\right)}{{sin}\left(\theta\right)}.{ln}\left(\mathrm{1}−{sin}^{\mathrm{2}} \left(\theta\right)\right){d}\theta \\ $$$$\:\:\:\overset{{sin}\left(\theta\right)={y}} {=}\:\int_{\mathrm{0}} ^{\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}−{y}^{\mathrm{2}} \right)}{{y}}{dy}=−\int_{\mathrm{0}} ^{\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{y}^{\mathrm{2}{n}−\mathrm{1}} }{{n}}{dy} \\ $$$$\:\:\:\:\:\:=−\Sigma\left[\frac{{y}^{\mathrm{2}{n}} }{\mathrm{2}{n}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =\frac{−\mathrm{1}}{\mathrm{2}}\:{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:=\frac{−\mathrm{1}}{\mathrm{2}}\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\right\}=\frac{−\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)… \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\boldsymbol{\phi}=\frac{−\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\:\:\:\:\:………\:\:\:\:\boldsymbol{\phi}\:=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{12}\pi}{\mathrm{48}}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{12}}{\mathrm{48}}\:{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\:….. \\ $$$$\:\:\:\:\:\:\:\:\:\:……..\boldsymbol{\phi}=\frac{\mathrm{1}}{\mathrm{48}}\left\{\mathrm{5}\pi^{\mathrm{2}} −\mathrm{12}\pi{ln}\left(\mathrm{2}\right)+\mathrm{12}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Dwaipayan Shikari last updated on 11/Apr/21
$${Nice}\:{sir}! \\ $$
Commented by mnjuly1970 last updated on 11/Apr/21
$$\:\:{thanks}\:{alot}\:{mr}\:{psaan}….. \\ $$