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Question Number 128244 by mnjuly1970 last updated on 05/Jan/21

               ...nice  calculus...       prove  that  ::Ω=  ∫_0 ^(π/4) ln(sin(x))d=((−π)/4)log(2)−(G/2)      log(2sin(x))=Σ_(n=1) ^∞ ((−1)/n)cos(2nx)   Ω= ∫_0 ^( (π/4)) {−log(2)−Σ_(n=1) ^∞ ((cos(2nx))/n)}dx  =((−π)/4)log(2)−Σ_(n=1) ^∞ ∫_0 ^( (π/4)) ((cos(2nx))/n)dx   =((−π)/4)log(2)−Σ_(n=1) ^∞ [(1/(2n^2 ))sin(2nx)]_0 ^(π/4)     =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ ((sin(((nπ)/2)))/n^2 )   =((−π)/4)log(2)−(1/2){(1/1^2 )−(1/3^2 ) +(1/5^2 )−..}   =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)^2 ))              ∴  Ω =((−π)/4)log(2)−(G/2)  ✓                       G:= catalan  constant...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:\:::\Omega=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left({x}\right)\right){d}=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{{G}}{\mathrm{2}} \\ $$$$\:\:\:\:{log}\left(\mathrm{2}{sin}\left({x}\right)\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{−\mathrm{1}}{{n}}{cos}\left(\mathrm{2}{nx}\right) \\ $$$$\:\Omega=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \left\{−{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}\right\}{dx} \\ $$$$=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}{dx} \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }{sin}\left(\mathrm{2}{nx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\:\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)}{{n}^{\mathrm{2}} } \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }−..\right\} \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\Omega\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{{G}}{\mathrm{2}}\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{G}:=\:{catalan}\:\:{constant}... \\ $$

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