advanced-calculus-prove-that-0-1-ln-x-cos-2-pix-dx-ln-2pi-4-1-8-m-n-1970-

Question Number 117280 by mnjuly1970 last updated on 10/Oct/20

... advanced calculus... prove that:: ∫_0 ^( 1) ln(Γ(x)).cos^2 (πx)dx =((ln(2π))/4)+(1/8) m.n.1970

$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}…\: \\ $$$$ \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right).{cos}^{\mathrm{2}} \left(\pi{x}\right){dx} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{ln}\left(\mathrm{2}\pi\right)}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$$$\: \\ $$

Answered by AbduraufKodiriy last updated on 10/Oct/20

∫_a ^( b) f(x)dx=∫_a ^( b) f(a+b−x)dx I=∫_0 ^( 1) ln(𝚪(x))cos^2 (πx)dx=∫_0 ^( 1) ln(𝚪(1−x))cos^2 (π(1−x))dx= =∫_0 ^( 1) ln(𝚪(1−x))cos^2 (πx)dx Therefore: 2I=∫_0 ^( 1) (ln(𝚪(x))+ln(𝚪(1−x)))cos^2 (πx)dx ⇒ ⇒ I=(1/2)∫_0 ^( 1) ln(𝚪(x)𝚪(1−x))cos^2 (πx)dx=(1/2)∫_0 ^( 1) ln((π/(sin(πx))))cos^2 (πx)dx= =((ln(π))/2)∫_0 ^( 1) cos^2 (πx)dx−(1/2)∫cos^2 (πx)ln(sin(πx))dx; I_1 =∫_0 ^( 1) cos^2 (πx)ln(sin(πx))dx ⇒ I=((ln(π))/4)−(1/2)I_1 I_1 = determinant (((πx=t ⇒ x=(t/π))),((dx=(1/π)dt)))=(1/π)∫_0 ^( π) cos^2 (t)ln(sin(t))dt= =(1/(2π))∫_0 ^( π) ln(sin(t))dt+(1/(2π))∫_0 ^( π) cos(2t)ln(sin(t))dt; We know: ∫_0 ^( π) ln(sin(t))dt=−πln2 I_1 =−(1/2)ln(2)+(1/2)∫_0 ^( π) cos(2t)ln(sin(t))dt=−(1/2)ln2+(((sin(2t)ln(sin(t)))/(4π))−(t/(4π))−((sin(2t))/(8π)))∣_0 ^π = =−(1/2)ln(2)−(1/4) ⇒ I=((ln(π))/4)+((ln(2))/4)+(1/8)=((ln(2π))/4)+(1/8)

$$\int_{\boldsymbol{{a}}} ^{\:\boldsymbol{{b}}} \boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\int_{\boldsymbol{{a}}} ^{\:\boldsymbol{{b}}} \boldsymbol{{f}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{x}}\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\boldsymbol{{x}}\right)\right)\boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\mathrm{1}−\boldsymbol{{x}}\right)\right)\boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\left(\mathrm{1}−\boldsymbol{{x}}\right)\right)\boldsymbol{{dx}}= \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\mathrm{1}−\boldsymbol{{x}}\right)\right)\boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\boldsymbol{{x}}\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{Therefore}}:\:\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\boldsymbol{{x}}\right)\right)+\boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\mathrm{1}−\boldsymbol{{x}}\right)\right)\right)\boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\boldsymbol{{x}}\right)\boldsymbol{{dx}}\:\Rightarrow \\ $$$$\Rightarrow\:\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\boldsymbol{{x}}\right)\boldsymbol{\Gamma}\left(\mathrm{1}−\boldsymbol{{x}}\right)\right)\boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\boldsymbol{{x}}\right)\boldsymbol{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{ln}}\left(\frac{\pi}{\boldsymbol{{sin}}\left(\pi\boldsymbol{{x}}\right)}\right)\boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\boldsymbol{{x}}\right)\boldsymbol{{dx}}= \\ $$$$=\frac{\boldsymbol{{ln}}\left(\pi\right)}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\boldsymbol{{x}}\right)\boldsymbol{{dx}}−\frac{\mathrm{1}}{\mathrm{2}}\int\boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\boldsymbol{{x}}\right)\boldsymbol{{ln}}\left(\boldsymbol{{sin}}\left(\pi\boldsymbol{{x}}\right)\right)\boldsymbol{{dx}}; \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} \boldsymbol{{cos}}^{\mathrm{2}} \left(\pi\boldsymbol{{x}}\right)\boldsymbol{{ln}}\left(\boldsymbol{{sin}}\left(\pi\boldsymbol{{x}}\right)\right)\boldsymbol{{dx}}\:\Rightarrow\:\boldsymbol{{I}}=\frac{\boldsymbol{{ln}}\left(\pi\right)}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{I}}_{\mathrm{1}} \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =\begin{vmatrix}{\pi\boldsymbol{{x}}=\boldsymbol{{t}}\:\Rightarrow\:\boldsymbol{{x}}=\frac{\boldsymbol{{t}}}{\pi}}\\{\boldsymbol{{dx}}=\frac{\mathrm{1}}{\pi}\boldsymbol{{dt}}}\end{vmatrix}=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\pi} \boldsymbol{{cos}}^{\mathrm{2}} \left(\boldsymbol{{t}}\right)\boldsymbol{{ln}}\left(\boldsymbol{{sin}}\left(\boldsymbol{{t}}\right)\right)\boldsymbol{{dt}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\:\pi} \boldsymbol{{ln}}\left(\boldsymbol{{sin}}\left(\boldsymbol{{t}}\right)\right)\boldsymbol{{dt}}+\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\:\pi} \boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{t}}\right)\boldsymbol{{ln}}\left(\boldsymbol{{sin}}\left(\boldsymbol{{t}}\right)\right)\boldsymbol{{dt}};\:\boldsymbol{{We}}\:\boldsymbol{{know}}:\:\int_{\mathrm{0}} ^{\:\pi} \boldsymbol{{ln}}\left(\boldsymbol{{sin}}\left(\boldsymbol{{t}}\right)\right)\boldsymbol{{dt}}=−\pi\boldsymbol{{ln}}\mathrm{2} \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{ln}}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi} \boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{t}}\right)\boldsymbol{{ln}}\left(\boldsymbol{{sin}}\left(\boldsymbol{{t}}\right)\right)\boldsymbol{{dt}}=−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{ln}}\mathrm{2}+\left(\frac{\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{t}}\right)\boldsymbol{{ln}}\left(\boldsymbol{{sin}}\left(\boldsymbol{{t}}\right)\right)}{\mathrm{4}\pi}−\frac{\boldsymbol{{t}}}{\mathrm{4}\pi}−\frac{\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{t}}\right)}{\mathrm{8}\pi}\right)\mid_{\mathrm{0}} ^{\pi} = \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{ln}}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:\boldsymbol{{I}}=\frac{\boldsymbol{{ln}}\left(\pi\right)}{\mathrm{4}}+\frac{\boldsymbol{{ln}}\left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\boldsymbol{{ln}}\left(\mathrm{2}\pi\right)}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Commented by AbduraufKodiriy last updated on 10/Oct/20