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Question-45498

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Question Number 45498 by Sanjarbek last updated on 13/Oct/18
Commented by Meritguide1234 last updated on 13/Oct/18
not solvable
$${not}\:{solvable} \\ $$
Commented by MJS last updated on 13/Oct/18
possible but not elementary. I′ll post it later
$$\mathrm{possible}\:\mathrm{but}\:\mathrm{not}\:\mathrm{elementary}.\:\mathrm{I}'\mathrm{ll}\:\mathrm{post}\:\mathrm{it}\:\mathrm{later} \\ $$
Commented by maxmathsup by imad last updated on 13/Oct/18
let f(t)=∫ ln(sin(tx))dx ⇒f^′ (t)= ∫ ((x cos(tx))/(sin(tx)))dx =_(tx =u) ∫ (u/t) ((cos(u))/(sin(u))) (du/t) =(1/t^2 ) ∫ u((cos(u))/(sin(u))) du also changement tan((u/2))=α give f(t)=(1/t^2 ) ∫ 2 arctan(α)(((1−α^2 )/(1+α^2 ))/((2α)/(1+α^2 ))) ((2dα)/(1+α^2 )) =(4/t^2 ) ∫ ((arctan(α))/(1+α^2 )) dα let then introduce the parametric function ϕ(u) =∫ ((arctan(uα))/(1+α^2 ))dα ⇒ϕ^′ (u) =∫ (α/((1+α^2 )(1+u^2 α^2 )))dα let decompose F(α)=((aα +b)/(1+α^2 )) +((cα +d)/(1+u^2 α^2 )) ....be continued...
$${let}\:{f}\left({t}\right)=\int\:{ln}\left({sin}\left({tx}\right)\right){dx}\:\Rightarrow{f}^{'} \left({t}\right)=\:\int\:\frac{{x}\:{cos}\left({tx}\right)}{{sin}\left({tx}\right)}{dx}\:\: \\ $$$$=_{{tx}\:={u}} \:\:\:\:\int\:\:\frac{{u}}{{t}}\:\frac{{cos}\left({u}\right)}{{sin}\left({u}\right)}\:\frac{{du}}{{t}}\:=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\int\:\:\:{u}\frac{{cos}\left({u}\right)}{{sin}\left({u}\right)}\:{du}\:\:\:{also}\:{changement}\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)=\alpha \\ $$$${give}\:{f}\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\int\:\:\mathrm{2}\:{arctan}\left(\alpha\right)\frac{\frac{\mathrm{1}−\alpha^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{2}} }}{\frac{\mathrm{2}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }}\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}}{{t}^{\mathrm{2}} }\:\int\:\:\:\:\frac{{arctan}\left(\alpha\right)}{\mathrm{1}+\alpha^{\mathrm{2}} }\:{d}\alpha\:{let}\:{then}\:{introduce}\:{the}\:{parametric}\:{function} \\ $$$$\varphi\left({u}\right)\:=\int\:\:\frac{{arctan}\left({u}\alpha\right)}{\mathrm{1}+\alpha^{\mathrm{2}} }{d}\alpha\:\Rightarrow\varphi^{'} \left({u}\right)\:=\int\:\:\:\frac{\alpha}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} \alpha^{\mathrm{2}} \right)}{d}\alpha\:{let}\:{decompose} \\ $$$${F}\left(\alpha\right)=\frac{{a}\alpha\:+{b}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:+\frac{{c}\alpha\:+{d}}{\mathrm{1}+{u}^{\mathrm{2}} \alpha^{\mathrm{2}} }\:….{be}\:{continued}… \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 14/Oct/18
∫ln sin x dx=∫ln (−(i/2)(e^(ix) −e^(−ix) )) dx= [t=ix → dx=−idt =−i∫ln (−(i/2)(e^t −e^(−t) )) dt= =−i∫ln (−(i/2)) dt−i∫ln (e^t −e^(−t) ) dt the first one is easy: −i∫ln (−(i/2)) dt=−i ln (−(i/2)) ∫dt=−i ln (−(i/2)) t= =ln (−(i/2)) x the second one: −i∫ln (e^t −e^(−t) ) dt= [((∫f′g=fg−∫fg′)),((f′=1 → f=t)),((g=ln (e^t −e^(−t) ) → g′=((e^t +e^(−t) )/(e^t −e^(−t) )))) ] =−itln (e^t −e^(−t) ) +i∫t((e^t +e^(−t) )/(e^t −e^(−t) ))dt the first term: −itln (e^t −e^(−t) )=xln (e^(ix) −e^(−ix) ) the second term: i∫t((e^t +e^(−t) )/(e^t −e^(−t) ))dt=i∫t((e^(2t) +1)/(e^(2t) −1))dt= [u=e^(2t) −1 → dt=(e^(−2t) /2)du] =(i/4)∫(((u+2)ln (u+1))/(u(u+1)))du= =(i/2)∫((ln (u+1))/u)du+(i/4)∫((ln (u+1))/(u+1))du the first one: (i/2)∫((ln (u+1))/u)du= [v=−u → du=−dv] =(i/2)∫((ln (1−v))/v)dv=−(i/2)∫−((ln (1−v))/v)dv= this is a special integral (dilogarithm) =−(i/2)Li_2 v=−(i/2)Li_2 (−u)=−(i/2)Li_2 (1−e^(2t) )= =−(i/2)Li_2 (1−e^(2ix) ) the second one: (i/4)∫((ln (u+1))/(u+1))du= [w=ln (u+1) → du=(u+1)dw] =(i/4)∫wdw=(i/8)w^2 =(i/8)ln^2 (u+1)=(i/8)ln^2 (e^(2t) )= =(i/2)t^2 =−(i/2)x^2 so we have ∫ln sin x dx= =xln (−(i/2)) +xln (e^(ix) −e^(−ix) ) −(i/2)Li_2 (1−e^(2ix) ) −(i/2)x^2 +C please check...
$$\int\mathrm{ln}\:\mathrm{sin}\:{x}\:{dx}=\int\mathrm{ln}\:\left(−\frac{\mathrm{i}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} \right)\right)\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{i}{x}\:\rightarrow\:{dx}=−\mathrm{i}{dt}\right. \\ $$$$=−\mathrm{i}\int\mathrm{ln}\:\left(−\frac{\mathrm{i}}{\mathrm{2}}\left(\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} \right)\right)\:{dt}= \\ $$$$=−\mathrm{i}\int\mathrm{ln}\:\left(−\frac{\mathrm{i}}{\mathrm{2}}\right)\:{dt}−\mathrm{i}\int\mathrm{ln}\:\left(\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} \right)\:{dt} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{one}\:\mathrm{is}\:\mathrm{easy}: \\ $$$$−\mathrm{i}\int\mathrm{ln}\:\left(−\frac{\mathrm{i}}{\mathrm{2}}\right)\:{dt}=−\mathrm{i}\:\mathrm{ln}\:\left(−\frac{\mathrm{i}}{\mathrm{2}}\right)\:\int{dt}=−\mathrm{i}\:\mathrm{ln}\:\left(−\frac{\mathrm{i}}{\mathrm{2}}\right)\:{t}= \\ $$$$=\mathrm{ln}\:\left(−\frac{\mathrm{i}}{\mathrm{2}}\right)\:{x} \\ $$$$\mathrm{the}\:\mathrm{second}\:\mathrm{one}: \\ $$$$−\mathrm{i}\int\mathrm{ln}\:\left(\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} \right)\:{dt}= \\ $$$$\:\:\:\:\:\begin{bmatrix}{\int{f}'{g}={fg}−\int{fg}'}\\{{f}'=\mathrm{1}\:\rightarrow\:{f}={t}}\\{{g}=\mathrm{ln}\:\left(\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} \right)\:\rightarrow\:{g}'=\frac{\mathrm{e}^{{t}} +\mathrm{e}^{−{t}} }{\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} }}\end{bmatrix} \\ $$$$=−\mathrm{i}{t}\mathrm{ln}\:\left(\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} \right)\:+\mathrm{i}\int{t}\frac{\mathrm{e}^{{t}} +\mathrm{e}^{−{t}} }{\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} }{dt} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{term}: \\ $$$$−\mathrm{i}{t}\mathrm{ln}\:\left(\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} \right)={x}\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} \right) \\ $$$$\mathrm{the}\:\mathrm{second}\:\mathrm{term}: \\ $$$$\mathrm{i}\int{t}\frac{\mathrm{e}^{{t}} +\mathrm{e}^{−{t}} }{\mathrm{e}^{{t}} −\mathrm{e}^{−{t}} }{dt}=\mathrm{i}\int{t}\frac{\mathrm{e}^{\mathrm{2}{t}} +\mathrm{1}}{\mathrm{e}^{\mathrm{2}{t}} −\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{e}^{\mathrm{2}{t}} −\mathrm{1}\:\rightarrow\:{dt}=\frac{\mathrm{e}^{−\mathrm{2}{t}} }{\mathrm{2}}{du}\right] \\ $$$$=\frac{\mathrm{i}}{\mathrm{4}}\int\frac{\left({u}+\mathrm{2}\right)\mathrm{ln}\:\left({u}+\mathrm{1}\right)}{{u}\left({u}+\mathrm{1}\right)}{du}= \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\left({u}+\mathrm{1}\right)}{{u}}{du}+\frac{\mathrm{i}}{\mathrm{4}}\int\frac{\mathrm{ln}\:\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}{du} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{one}: \\ $$$$\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\left({u}+\mathrm{1}\right)}{{u}}{du}= \\ $$$$\:\:\:\:\:\left[{v}=−{u}\:\rightarrow\:{du}=−{dv}\right] \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\int\frac{\mathrm{ln}\:\left(\mathrm{1}−{v}\right)}{{v}}{dv}=−\frac{\mathrm{i}}{\mathrm{2}}\int−\frac{\mathrm{ln}\:\left(\mathrm{1}−{v}\right)}{{v}}{dv}= \\ $$$$\:\:\:\:\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{special}\:\mathrm{integral}\:\left(\mathrm{dilogarithm}\right) \\ $$$$=−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \:{v}=−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \:\left(−{u}\right)=−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{1}−\mathrm{e}^{\mathrm{2}{t}} \right)= \\ $$$$=−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{1}−\mathrm{e}^{\mathrm{2i}{x}} \right) \\ $$$$\mathrm{the}\:\mathrm{second}\:\mathrm{one}: \\ $$$$\frac{\mathrm{i}}{\mathrm{4}}\int\frac{\mathrm{ln}\:\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}{du}= \\ $$$$\:\:\:\:\:\left[{w}=\mathrm{ln}\:\left({u}+\mathrm{1}\right)\:\rightarrow\:{du}=\left({u}+\mathrm{1}\right){dw}\right] \\ $$$$=\frac{\mathrm{i}}{\mathrm{4}}\int{wdw}=\frac{\mathrm{i}}{\mathrm{8}}{w}^{\mathrm{2}} =\frac{\mathrm{i}}{\mathrm{8}}\mathrm{ln}^{\mathrm{2}} \:\left({u}+\mathrm{1}\right)=\frac{\mathrm{i}}{\mathrm{8}}\mathrm{ln}^{\mathrm{2}} \:\left(\mathrm{e}^{\mathrm{2}{t}} \right)= \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}{t}^{\mathrm{2}} =−\frac{\mathrm{i}}{\mathrm{2}}{x}^{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int\mathrm{ln}\:\mathrm{sin}\:{x}\:{dx}= \\ $$$$={x}\mathrm{ln}\:\left(−\frac{\mathrm{i}}{\mathrm{2}}\right)\:+{x}\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} \right)\:−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{1}−\mathrm{e}^{\mathrm{2i}{x}} \right)\:−\frac{\mathrm{i}}{\mathrm{2}}{x}^{\mathrm{2}} +{C} \\ $$$$\mathrm{please}\:\mathrm{check}… \\ $$

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