Question Number 123745 by mnjuly1970 last updated on 27/Nov/20
![...advanced calculus... evaluation of: Ω=∫_0 ^( (π/2)) sin(x)log(sin(x))dx by using the euler beta and gamma function: β(p,(1/2))=2∫_0 ^(π/2) sin^(2p−1) (x)dx ((dβ(p,(1/2)))/dp) =2∫_0 ^(π/2) 2sin^(2p−1) (x)ln(sin(x))dx =4∫_0 ^(π/2) sin^(2p−1) (x)ln(sin(x))dx Ω=(1/4)[((dβ(p,(1/2)))/dp)]_(p=1) ((d(β(p,(1/2))))/dp)=(√π)[((Γ′(p)Γ(p+(1/2))−Γ′(p+(1/2))Γ(p))/(Γ^2 (p+(1/2))))]_(p=1) Ω=(1/4)((√π)[((Γ^′ (1)Γ((3/2))−Γ′((3/2))Γ(1))/(Γ^2 ((3/2))))]) =((√π)/4)[((−γ(((√π)/2))−((√π)/2)(2−γ−2ln(2)))/(π/4))] =((√π)/4)∗((√π)/2)(((−γ−2+γ+2ln(2))/(π/4))) =ln(2)−1 ... m.n.july.1970...](Q123745.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{advanced}\:\:\:{calculus}... \\ $$$$\:\:{evaluation}\:\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}\left({x}\right){log}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:{by}\:{using}\:{the}\:{euler}\:{beta}\:{and}\:{gamma}\:{function}: \\ $$$$\:\:\:\:\:\:\:\:\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){dx} \\ $$$$\:\:\:\frac{{d}\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)}{{dp}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{d}\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)}{{dp}}\right]_{{p}=\mathrm{1}} \\ $$$$\:\frac{{d}\left(\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)\right)}{{dp}}=\sqrt{\pi}\left[\frac{\Gamma'\left({p}\right)\Gamma\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\Gamma'\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({p}\right)}{\Gamma^{\mathrm{2}} \left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\right]_{{p}=\mathrm{1}} \\ $$$$\:\Omega=\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\pi}\left[\frac{\Gamma^{'} \left(\mathrm{1}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\right]\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}}\left[\frac{−\gamma\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right)−\frac{\sqrt{\pi}}{\mathrm{2}}\left(\mathrm{2}−\gamma−\mathrm{2}{ln}\left(\mathrm{2}\right)\right)}{\frac{\pi}{\mathrm{4}}}\right] \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}}\ast\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{−\gamma−\mathrm{2}+\gamma+\mathrm{2}{ln}\left(\mathrm{2}\right)}{\frac{\pi}{\mathrm{4}}}\right) \\ $$$$={ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:...\:{m}.{n}.{july}.\mathrm{1970}... \\ $$$$\:\:\: \\ $$