Question Number 118705 by Lordose last updated on 19/Oct/20
$$ \\ $$$$ \\ $$$$…\:\blacklozenge\mathrm{Advanced}\:\mathrm{Calculus}\blacklozenge… \\ $$$$ \\ $$$$\mathrm{Evaluate}:: \\ $$$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\:\mathrm{1}\:} \frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$ \\ $$$$…\spadesuit\boldsymbol{\mathrm{L}\phi\mathrm{rD}}\:\boldsymbol{\varnothing\mathrm{sE}}\spadesuit… \\ $$$$ \\ $$$$…\clubsuit\boldsymbol{\mathrm{GooD}}\:\boldsymbol{\mathrm{LucK}}\clubsuit \\ $$
Answered by mathdave last updated on 19/Oct/20
$${solution} \\ $$$${let}\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:\left({put}\:{y}=\mathrm{sin}^{−\mathrm{1}} {x}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{y}\mathrm{cos}{y}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {y}}{dy}\:\:\left({using}\:{IBP}\right) \\ $$$${I}=\left({y}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}{y}\right)\right)_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}{y}\right){dy} \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}{y}\right){dy} \\ $$$${according}\:{to}\:{legendary}\:{chi}−{function} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left({r}\mathrm{sin}\theta\right){d}\theta=\mathrm{2}\chi_{\mathrm{2}} \left(\frac{\sqrt{\mathrm{1}+{r}^{\mathrm{2}} }−\mathrm{1}}{{r}}\right) \\ $$$${if}\:{r}=\mathrm{1}\:\:{then} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}{y}\right){dy}=\mathrm{2}\chi_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\because{note}\:{that}\:\chi_{{v}} \left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({Li}_{{v}} \left({z}\right)−{Li}_{{v}} \left(−{z}\right)\right) \\ $$$${then} \\ $$$$\chi_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({Li}_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\right) \\ $$$$\because\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}{y}\right){dy}=\mathrm{2}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\left({Li}_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\right) \\ $$$$\because{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\left({Li}_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\right) \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\left({Li}_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}−\sqrt{\mathrm{2}}\right)=\mathrm{0}.\mathrm{38841}\right. \\ $$$${OR}\: \\ $$$$\:{legendary}\:{chi}−{function}\:{of} \\ $$$$\mathrm{2}\chi_{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\mathrm{2}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{4}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\because\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}{y}\right){dy}=\left(\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$$\because{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\left(\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}}\right)=\frac{\mathrm{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{0}.\mathrm{38841} \\ $$$${the}\:{two}\:{answer}\:{are}\:{correct}\: \\ $$$${by}\:{mathdave}\left(\mathrm{19}/\mathrm{10}/\mathrm{2020}\right) \\ $$
Commented by Lordose last updated on 19/Oct/20
$$\mathrm{Nice}\:\mathrm{one}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by mindispower last updated on 19/Oct/20
![x=sin(t) Ω=∫_0 ^(π/2) ((tcos(t))/(1+sin^2 (t)))=[tarctan(sin(t))]−∫_0 ^(π/2) arctan(sin(t))dt =(π^2 /8)−∫_0 ^1 ((arctan(r))/( (√(1−r^2 ))))dr f(s)=∫_0 ^1 ((arctan(sr))/( (√(1−r^2 ))))dr,f(0)=0 f′(s)=∫_0 ^1 (r/( (√(1−r^2 )) (1+s^2 r^2 )))dr,r=sin(x) f′(s)=∫_0 ^(π/2) ((sin(x))/(1+s^2 sin^2 (x)))=∫_0 ^(π/2) ((sin(x))/(1+s^2 −s^2 cos^2 (x))) =−(1/( s(√(1+s^2 ))))∫_0 ((s(−sin(x))dx)/( (√(1+s^2 )) (1−((s/( (√(1+s^2 ))))cos(x))^2 )) =−(1/(s(√(1+s^2 ))))[arcth((s/( (√(1+s^2 ))))cos(x))]_0 ^(π/2) =(1/(s(√(1+s^2 ))))arcth((s/( (√(1+s^2 )))))=f′(s) f(s)=∫_0 ^s (1/( s(√(1+s^2 ))))arcth((s/( (√(1+s^2 ))))) we want f(1) f(1)=∫_0 ^1 (1/(s(√(1+s^2 ))))arcth((s/( (√(1+s^2 ))))) let s=sh(t)⇒f(1)=∫_0 ^(sh^− (1)) ((arcth(((sh(t))/(ch(t))))ch(t)dt)/(sh(t)ch(t))) =∫_0 ^(sh^− (1)) (t/(sh(t)))dt...by part ∫(1/(sh(t)))=ln(1−e^(−t) )−ln(1+e^(−t) ) f(1)=[tln(((1−e^(−t) )/(1+e^(−t) )))]_0 ^(sh^− (t)) −∫_0 ^(sh^− (1)) ln(1−e^(−t) )dt +∫_0 ^(sh^− (t)) ln(1+e^(−t) ) ∫_0 ^(sh^− (1)) ln(1−e^(−t) )dt,e^(−t) =w =∫_1 ^e^(−sh^− (1)) ((ln(1−w))/(−w))dw=∫_0 ^1 ((ln(1−w))/w)−∫_0 ^e^(−sh^− (1)) ((ln(1−w))/w)dw =Li_2 (e^(−sh^− (1)) )−Li_2 (1) ∫_0 ^e^(−sh^− (1)) ln(1+e^(−t) )dt,e^(−t) =−w ⇒∫_(−1) ^(−e^(−sh^− (1)) ) ((ln(1−w)dw)/(−w))=Li_2 (−e^(−sh^− (1)) )−li_2 (−1) we get,a=sh^− (1) f(1)=aln(((1−e^(−a) )/(1+e^(−a) )))+Li_2 (−e^(−a) )−li_2 (−1)−Li_2 (e^(−a) )+Li_2 (1) Li_2 (1)=ζ(1)=(π^2 /6),li_2 (−1)=−(π^2 /(12)) f(1)=(π^2 /4)+Li_2 (−e^(−a) )−Li_2 (e^(−a) )+aln(((1−e^(−a) )/(1+e^(−a) ))) Ω=(π^2 /8)−f(1).. i will try if there is possibility to give Li_2 (...)−Li_2 (...) by elementry function](https://www.tinkutara.com/question/Q118782.png)
$${x}={sin}\left({t}\right) \\ $$$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tcos}\left({t}\right)}{\mathrm{1}+{sin}^{\mathrm{2}} \left({t}\right)}=\left[{tarctan}\left({sin}\left({t}\right)\right)\right]−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {arctan}\left({sin}\left({t}\right)\right){dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctan}\left({r}\right)}{\:\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }}{dr} \\ $$$${f}\left({s}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctan}\left({sr}\right)}{\:\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }}{dr},{f}\left(\mathrm{0}\right)=\mathrm{0}\:\: \\ $$$${f}'\left({s}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{r}}{\:\sqrt{\mathrm{1}−{r}^{\mathrm{2}} \:}\:\left(\mathrm{1}+{s}^{\mathrm{2}} {r}^{\mathrm{2}} \right)}{dr},{r}={sin}\left({x}\right) \\ $$$${f}'\left({s}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left({x}\right)}{\mathrm{1}+{s}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left({x}\right)}{\mathrm{1}+{s}^{\mathrm{2}} −{s}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)} \\ $$$$=−\frac{\mathrm{1}}{\:{s}\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}\int_{\mathrm{0}} \frac{{s}\left(−{sin}\left({x}\right)\right){dx}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }\:\left(\mathrm{1}−\left(\frac{{s}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}{cos}\left({x}\right)\right)^{\mathrm{2}} \right.} \\ $$$$=−\frac{\mathrm{1}}{{s}\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}\left[{arcth}\left(\frac{{s}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}{cos}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{{s}\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}{arcth}\left(\frac{{s}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}\right)={f}'\left({s}\right) \\ $$$${f}\left({s}\right)=\int_{\mathrm{0}} ^{{s}} \frac{\mathrm{1}}{\:{s}\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}{arcth}\left(\frac{{s}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}\right) \\ $$$${we}\:{want}\:{f}\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{s}\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}{arcth}\left(\frac{{s}}{\:\sqrt{\mathrm{1}+{s}^{\mathrm{2}} }}\right) \\ $$$${let}\:{s}={sh}\left({t}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{{sh}^{−} \left(\mathrm{1}\right)} \frac{{arcth}\left(\frac{{sh}\left({t}\right)}{{ch}\left({t}\right)}\right){ch}\left({t}\right){dt}}{{sh}\left({t}\right){ch}\left({t}\right)} \\ $$$$=\int_{\mathrm{0}} ^{{sh}^{−} \left(\mathrm{1}\right)} \frac{{t}}{{sh}\left({t}\right)}{dt}…{by}\:{part} \\ $$$$\int\frac{\mathrm{1}}{{sh}\left({t}\right)}={ln}\left(\mathrm{1}−{e}^{−{t}} \right)−{ln}\left(\mathrm{1}+{e}^{−{t}} \right) \\ $$$${f}\left(\mathrm{1}\right)=\left[{tln}\left(\frac{\mathrm{1}−{e}^{−{t}} }{\mathrm{1}+{e}^{−{t}} }\right)\right]_{\mathrm{0}} ^{{sh}^{−} \left({t}\right)} −\int_{\mathrm{0}} ^{{sh}^{−} \left(\mathrm{1}\right)} {ln}\left(\mathrm{1}−{e}^{−{t}} \right){dt} \\ $$$$+\int_{\mathrm{0}} ^{{sh}^{−} \left({t}\right)} {ln}\left(\mathrm{1}+{e}^{−{t}} \right) \\ $$$$\int_{\mathrm{0}} ^{{sh}^{−} \left(\mathrm{1}\right)} {ln}\left(\mathrm{1}−{e}^{−{t}} \right){dt},{e}^{−{t}} ={w} \\ $$$$=\int_{\mathrm{1}} ^{{e}^{−{sh}^{−} \left(\mathrm{1}\right)} } \frac{{ln}\left(\mathrm{1}−{w}\right)}{−{w}}{dw}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{w}\right)}{{w}}−\int_{\mathrm{0}} ^{{e}^{−{sh}^{−} \left(\mathrm{1}\right)} } \frac{{ln}\left(\mathrm{1}−{w}\right)}{{w}}{dw} \\ $$$$={Li}_{\mathrm{2}} \left({e}^{−{sh}^{−} \left(\mathrm{1}\right)} \right)−\boldsymbol{{L}}{i}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{{e}^{−{sh}^{−} \left(\mathrm{1}\right)} } {ln}\left(\mathrm{1}+{e}^{−{t}} \right){dt},{e}^{−{t}} =−{w} \\ $$$$\Rightarrow\int_{−\mathrm{1}} ^{−{e}^{−{sh}^{−} \left(\mathrm{1}\right)} } \frac{{ln}\left(\mathrm{1}−{w}\right){dw}}{−{w}}={Li}_{\mathrm{2}} \left(−{e}^{−{sh}^{−} \left(\mathrm{1}\right)} \right)−{li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$${we}\:{get},{a}={sh}^{−} \left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{1}\right)={aln}\left(\frac{\mathrm{1}−{e}^{−{a}} }{\mathrm{1}+{e}^{−{a}} }\right)+{Li}_{\mathrm{2}} \left(−{e}^{−{a}} \right)−{li}_{\mathrm{2}} \left(−\mathrm{1}\right)−{Li}_{\mathrm{2}} \left({e}^{−{a}} \right)+{Li}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$${Li}_{\mathrm{2}} \left(\mathrm{1}\right)=\zeta\left(\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}},{li}_{\mathrm{2}} \left(−\mathrm{1}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${f}\left(\mathrm{1}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+{Li}_{\mathrm{2}} \left(−{e}^{−{a}} \right)−{Li}_{\mathrm{2}} \left({e}^{−{a}} \right)+{aln}\left(\frac{\mathrm{1}−{e}^{−{a}} }{\mathrm{1}+{e}^{−{a}} }\right) \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−{f}\left(\mathrm{1}\right)..\:{i}\:{will}\:{try}\:{if}\:{there}\:{is}\:{possibility} \\ $$$${to}\:{give}\:{Li}_{\mathrm{2}} \left(…\right)−{Li}_{\mathrm{2}} \left(…\right)\:{by}\:{elementry}\:{function} \\ $$$$ \\ $$