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Question Number 146775 by Willson last updated on 15/Jul/21

Prove that ∫^( 𝛑) _( 0) tln(sint)dt= −(𝛑^2 /2)ln(2)

$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\:\:\underset{\:\mathrm{0}} {\int}^{\:\boldsymbol{\pi}} \boldsymbol{{tln}}\left(\boldsymbol{{sint}}\right)\boldsymbol{{dt}}=\:−\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{2}}\boldsymbol{{ln}}\left(\mathrm{2}\right) \\ $$

Answered by ArielVyny last updated on 15/Jul/21

∫_0 ^π t(−ln2−Σ_(n≥1) ((cos(2nt))/n))dt −ln(2)∫_0 ^π tdt−∫_0 ^π (tΣ_(n≥1) ((cos(2nt))/n))dt −(π^2 /2)ln2 according that ∫_0 ^π tΣ_(n≥1) ((cos(2nt))/n)dt=0 ∫_0 ^π tln(sint)dt=−((π^2 ln2)/2)

$$\int_{\mathrm{0}} ^{\pi} {t}\left(−{ln}\mathrm{2}−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left(\mathrm{2}{nt}\right)}{{n}}\right){dt} \\ $$$$−{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\pi} {tdt}−\int_{\mathrm{0}} ^{\pi} \left({t}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left(\mathrm{2}{nt}\right)}{{n}}\right){dt} \\ $$$$−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}{ln}\mathrm{2}\:\:{according}\:{that}\:\int_{\mathrm{0}} ^{\pi} {t}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left(\mathrm{2}{nt}\right)}{{n}}{dt}=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\pi} {tln}\left({sint}\right){dt}=−\frac{\pi^{\mathrm{2}} {ln}\mathrm{2}}{\mathrm{2}} \\ $$$$ \\ $$

Answered by Olaf_Thorendsen last updated on 15/Jul/21

Ω = ∫_0 ^π tln(sint) dt (1) Let u = π−t Ω = ∫_0 ^π (π−u)ln(sinu) du (2) (((1)+(2))/2) : Ω = (π/2)∫_0 ^π ln(sint)dt (3) Ω = (π/2)∫_0 ^(π/2) ln(sint)dt+(π/2)∫_(π/2) ^π ln(sint)dt Let u = t−(π/2) Ω = (π/2)∫_0 ^(π/2) ln(sint)dt+(π/2)∫_0 ^(π/2) ln(cosu)du (4) Let t = 2u (3) : Ω = π∫_0 ^(π/2) ln(sin(2u)) du Ω = π∫_0 ^(π/2) ln2+ln(sinu)+ln(cosu) du (5) (5)−2×(4)w : −Ω = π∫_0 ^(π/2) ln2 du = (π^2 /2)ln(2) ⇒ Ω = −(π^2 /2)ln(2)

$$\Omega\:=\:\int_{\mathrm{0}} ^{\pi} {t}\mathrm{ln}\left(\mathrm{sin}{t}\right)\:{dt}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{u}\:=\:\pi−{t} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\pi} \left(\pi−{u}\right)\mathrm{ln}\left(\mathrm{sin}{u}\right)\:{du}\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{1}\right)+\left(\mathrm{2}\right)}{\mathrm{2}}\::\:\Omega\:=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}{t}\right){dt}\:\:\:\left(\mathrm{3}\right) \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{t}\right){dt}+\frac{\pi}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}{t}\right){dt} \\ $$$$\mathrm{Let}\:{u}\:=\:{t}−\frac{\pi}{\mathrm{2}} \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{t}\right){dt}+\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}{u}\right){du}\:\left(\mathrm{4}\right) \\ $$$$ \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{2}{u} \\ $$$$\left(\mathrm{3}\right)\::\:\Omega\:=\:\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{2}{u}\right)\right)\:{du} \\ $$$$\Omega\:=\:\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln2}+\mathrm{ln}\left(\mathrm{sin}{u}\right)+\mathrm{ln}\left(\mathrm{cos}{u}\right)\:{du}\:\:\left(\mathrm{5}\right) \\ $$$$ \\ $$$$\left(\mathrm{5}\right)−\mathrm{2}×\left(\mathrm{4}\right){w}\:: \\ $$$$−\Omega\:=\:\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln2}\:{du}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:\Omega\:=\:−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$

Answered by mathmax by abdo last updated on 16/Jul/21

Υ=∫_0 ^π t log(sint)dt ⇒Υ=∫_0 ^(π/2) tlog(sint)dt +∫_(π/2) ^π tlog(sint)dt(→t=(π/2)+z) =∫_0 ^(π/2) tlog(sint)dt[+∫_0 ^(π/2) ((π/2)+z)log(cost)dt =(π/2)∫_0 ^(π/2) log(cost)dt +∫_0 ^(π/2) t(log(sint)+log(cost))dt =−(π^2 /4)log2 +∫_0 ^(π/2) tlog(((sin(2t))/2))dt =−(π^2 /4)log2 +∫_0 ^(π/2) tlog(sin(2t)dt(→2t=u)−log2∫_0 ^(π/2) tdt =−(π^2 /4)log2 +∫_0 ^π (u/2)log(sinu)(du/2)−log2×(π^2 /8) =−((3π^2 )/8)log2 +(Υ/4) ⇒(1−(1/4))Υ=−((3π^2 )/8)log2 ⇒ (3/4)Υ=−((3π^2 )/8)log2 ⇒Υ=−(π^2 /2)log2

$$\Upsilon=\int_{\mathrm{0}} ^{\pi} \:\mathrm{t}\:\mathrm{log}\left(\mathrm{sint}\right)\mathrm{dt}\:\:\Rightarrow\Upsilon=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{tlog}\left(\mathrm{sint}\right)\mathrm{dt}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:\mathrm{tlog}\left(\mathrm{sint}\right)\mathrm{dt}\left(\rightarrow\mathrm{t}=\frac{\pi}{\mathrm{2}}+\mathrm{z}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{tlog}\left(\mathrm{sint}\right)\mathrm{dt}\left[+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\frac{\pi}{\mathrm{2}}+\mathrm{z}\right)\mathrm{log}\left(\mathrm{cost}\right)\mathrm{dt}\right. \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{log}\left(\mathrm{cost}\right)\mathrm{dt}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{t}\left(\mathrm{log}\left(\mathrm{sint}\right)+\mathrm{log}\left(\mathrm{cost}\right)\right)\mathrm{dt} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{log2}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{tlog}\left(\frac{\mathrm{sin}\left(\mathrm{2t}\right)}{\mathrm{2}}\right)\mathrm{dt} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{log2}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{tlog}\left(\mathrm{sin}\left(\mathrm{2t}\right)\mathrm{dt}\left(\rightarrow\mathrm{2t}=\mathrm{u}\right)−\mathrm{log2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{tdt}\right. \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{log2}\:+\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{u}}{\mathrm{2}}\mathrm{log}\left(\mathrm{sinu}\right)\frac{\mathrm{du}}{\mathrm{2}}−\mathrm{log2}×\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$=−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{log2}\:+\frac{\Upsilon}{\mathrm{4}}\:\Rightarrow\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\Upsilon=−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{log2}\:\Rightarrow \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\Upsilon=−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{log2}\:\Rightarrow\Upsilon=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\mathrm{log2} \\ $$$$ \\ $$

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