prove-that-I-0-pi-2-ln-1-sin-2-d-2G-pi-ln-2-G-catalan-constant-

Question Number 161089 by mnjuly1970 last updated on 12/Dec/21

prove that I= ∫_0 ^( (π/2)) ln ( 1+ sin (2 α )) dα = 2G − π ln ((√2) ) G: catalan constant

$$ \\ $$$$\:\:{prove}\:{that} \\ $$$$\:\:\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\left(\:\mathrm{1}+\:{sin}\:\left(\mathrm{2}\:\alpha\:\right)\right)\:{d}\alpha\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{2G}\:−\:\pi\:\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{G}:\:\:{catalan}\:{constant} \\ $$

Answered by Ar Brandon last updated on 11/Dec/21

I=∫_0 ^(π/2) ln(1+sin2α)dα =∫_0 ^(π/2) ln(cosα+sinα)^2 dα =2∫_0 ^(π/2) ln(cosα+sinα)dα =2(G−((πln2)/4))=2G−((πln2)/2) =2G−πln(√2)

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{sin2}\alpha\right){d}\alpha \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}\alpha+\mathrm{sin}\alpha\right)^{\mathrm{2}} {d}\alpha \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}\alpha+\mathrm{sin}\alpha\right){d}\alpha \\ $$$$\:\:\:=\mathrm{2}\left({G}−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\right)=\mathrm{2}{G}−\frac{\pi\mathrm{ln2}}{\mathrm{2}} \\ $$$$\:\:\:=\mathrm{2}{G}−\pi\mathrm{ln}\sqrt{\mathrm{2}} \\ $$

Answered by mnjuly1970 last updated on 12/Dec/21