Question Number 190692 by mnjuly1970 last updated on 09/Apr/23
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{sin}^{\:\mathrm{3}} \left({x}\:\right)\:\mathrm{ln}\left(\:{x}\:\right)}{{x}}\:\mathrm{d}{x}\:=\:?\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:@\:\mathrm{nice}\:β\:\mathrm{mathematics}\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by 07049753053 last updated on 09/Apr/23
![we know that sin^3 (x)=(1/4)(3sin(x)βsin(3x)) (3/4)β«_0 ^β ((sin(x)ln(x))/x)dxβ(1/4)β«_0 ^β ((sin(3x)ln(x))/x)dx (3/4)(((βππ)/2))β(1/4)β«_0 ^β ((sin(3x)ln(x))/x)dx β(3/8)ππβ(1/4)(d/da)β£_(a=1) β«_0 ^β ((sin(zx)x^a )/x)dx (d/da)β£_(a=1) β«_0 ^β sin(zx)x^(aβ1) dx let zx=u dx=(du/z) (d/da)β£_(a=1) (1/z)β«_0 ^β sin(u)((u/z))^(aβ1) du=(d/da)β£_(a=1) [(1/z^a )β«_0 ^β sin(u)u^(aβ1) du] from euler formula sin(x)=Im(e^(βix) ) (d/da)β£_(a=1) [(1/z^a )Imβ«_0 ^β e^(βiu) u^(aβ1) du] let ui=k du=(dk/i) (d/da)β£_(a=1) [(1/z^a )Imβ«_0 ^β e^(βk) ((k/i))^(aβ1) (dk/i)] (d/da)β£_(a=1) [(1/z^a )Im((1/i))πͺ(a)]=(d/da)β£_(a=1) [((βπͺ(a)sin(((πa)/2)))/z^a )] [β(z^(βa) /2)πͺ(a)(2πsin(((πa)/2))(log(z)βπ(a))βπcos(((πa)/2)))]_(a=1) β(z^(β1) /2)[2π(log(z)+π)=(π/z)log(z)+(π/z)π here z=3 (β(1/4))((π/3)log(3)+(π/3)π)β(3/8)Οπ=β(π/(12))log(3)β((ππ)/(12))β((3π)/8)Ξ³=(π/4)(((log(3))/3)β(π/3)β((3π)/4))=(π/4)(((log(3))/3)β((7π)/(12)))=(π/(12))(log(3)β((7π)/4))](https://www.tinkutara.com/question/Q190722.png)
$${we}\:{know}\:{that}\:{sin}^{\mathrm{3}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{sin}\left({x}\right)β{sin}\left(\mathrm{3}{x}\right)\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}β\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{β\boldsymbol{\gamma\pi}}{\mathrm{2}}\right)β\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$$$β\frac{\mathrm{3}}{\mathrm{8}}\boldsymbol{\gamma\pi}β\frac{\mathrm{1}}{\mathrm{4}}\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{zx}}\right)\boldsymbol{{x}}^{\boldsymbol{{a}}} }{\boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{zx}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}β\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{zx}}=\boldsymbol{\mathrm{u}}\:\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{z}}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{u}}\right)\left(\frac{\boldsymbol{\mathrm{u}}}{\boldsymbol{\mathrm{z}}}\right)^{\boldsymbol{\mathrm{a}}β\mathrm{1}} \boldsymbol{\mathrm{du}}=\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{\mathrm{1}}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\int_{\mathrm{0}} ^{\infty} \boldsymbol{{sin}}\left(\boldsymbol{{u}}\right)\boldsymbol{{u}}^{\boldsymbol{{a}}β\mathrm{1}} \boldsymbol{{du}}\right] \\ $$$$\boldsymbol{{from}}\:\boldsymbol{{euler}}\:\boldsymbol{{formula}} \\ $$$$\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{Im}}\left(\boldsymbol{{e}}^{β\boldsymbol{{ix}}} \right) \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{\mathrm{1}}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\boldsymbol{{Im}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{β\boldsymbol{{iu}}} \boldsymbol{{u}}^{\boldsymbol{{a}}β\mathrm{1}} \boldsymbol{{du}}\right] \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{ui}}=\boldsymbol{{k}}\:\boldsymbol{{du}}=\frac{\boldsymbol{{dk}}}{\boldsymbol{{i}}} \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{\mathrm{1}}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\boldsymbol{{Im}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{β\boldsymbol{{k}}} \left(\frac{\boldsymbol{{k}}}{\boldsymbol{{i}}}\right)^{\boldsymbol{{a}}β\mathrm{1}} \frac{\boldsymbol{{dk}}}{\boldsymbol{{i}}}\right] \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{\mathrm{1}}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\boldsymbol{{Im}}\left(\frac{\mathrm{1}}{\boldsymbol{{i}}}\right)\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\right]=\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{β\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\right] \\ $$$$\left[β\frac{{z}^{β{a}} }{\mathrm{2}}\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\left(\mathrm{2}\boldsymbol{\pi{sin}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)\left(\boldsymbol{{log}}\left(\boldsymbol{{z}}\right)β\boldsymbol{\psi}\left(\boldsymbol{{a}}\right)\right)β\boldsymbol{\pi{cos}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)\right)\right]_{\boldsymbol{{a}}=\mathrm{1}} \\ $$$$β\frac{\boldsymbol{{z}}^{β\mathrm{1}} }{\mathrm{2}}\left[\mathrm{2}\boldsymbol{\pi}\left(\boldsymbol{{log}}\left(\boldsymbol{{z}}\right)+\boldsymbol{\gamma}\right)=\frac{\boldsymbol{\pi}}{\boldsymbol{{z}}}\boldsymbol{{log}}\left(\boldsymbol{{z}}\right)+\frac{\boldsymbol{\gamma}}{\boldsymbol{{z}}}\boldsymbol{\pi}\:\right. \\ $$$$\boldsymbol{{here}}\:\boldsymbol{{z}}=\mathrm{3} \\ $$$$\left(β\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\boldsymbol{\pi}}{\mathrm{3}}\boldsymbol{{log}}\left(\mathrm{3}\right)+\frac{\boldsymbol{\gamma}}{\mathrm{3}}\boldsymbol{\pi}\right)β\frac{\mathrm{3}}{\mathrm{8}}\pi\boldsymbol{\gamma}=β\frac{\boldsymbol{\pi}}{\mathrm{12}}\boldsymbol{{log}}\left(\mathrm{3}\right)β\frac{\boldsymbol{\gamma\pi}}{\mathrm{12}}β\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{8}}\gamma=\frac{\boldsymbol{\pi}}{\mathrm{4}}\left(\frac{\boldsymbol{{log}}\left(\mathrm{3}\right)}{\mathrm{3}}β\frac{\boldsymbol{\gamma}}{\mathrm{3}}β\frac{\mathrm{3}\boldsymbol{\gamma}}{\mathrm{4}}\right)=\frac{\boldsymbol{\pi}}{\mathrm{4}}\left(\frac{\boldsymbol{{log}}\left(\mathrm{3}\right)}{\mathrm{3}}β\frac{\mathrm{7}\boldsymbol{\gamma}}{\mathrm{12}}\right)=\frac{\boldsymbol{\pi}}{\mathrm{12}}\left(\boldsymbol{{log}}\left(\mathrm{3}\right)β\frac{\mathrm{7}\boldsymbol{\gamma}}{\mathrm{4}}\right) \\ $$$$ \\ $$