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Question Number 190692 by mnjuly1970 last updated on 09/Apr/23
             calculate                      𝛗 = ∫_0 ^( ∞) (( sin^( 3) (x ) ln( x ))/x) dx = ?                                     @ nice βˆ’ mathematics
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{sin}^{\:\mathrm{3}} \left({x}\:\right)\:\mathrm{ln}\left(\:{x}\:\right)}{{x}}\:\mathrm{d}{x}\:=\:?\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:@\:\mathrm{nice}\:βˆ’\:\mathrm{mathematics}\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by 07049753053 last updated on 09/Apr/23
we know that sin^3 (x)=(1/4)(3sin(x)βˆ’sin(3x))  (3/4)∫_0 ^∞ ((sin(x)ln(x))/x)dxβˆ’(1/4)∫_0 ^∞ ((sin(3x)ln(x))/x)dx  (3/4)(((βˆ’π›„π›‘)/2))βˆ’(1/4)∫_0 ^∞ ((sin(3x)ln(x))/x)dx  βˆ’(3/8)π›„π›‘βˆ’(1/4)(d/da)∣_(a=1) ∫_0 ^∞ ((sin(zx)x^a )/x)dx  (d/da)∣_(a=1) ∫_0 ^∞ sin(zx)x^(aβˆ’1) dx  let zx=u dx=(du/z)  (d/da)∣_(a=1) (1/z)∫_0 ^∞ sin(u)((u/z))^(aβˆ’1) du=(d/da)∣_(a=1) [(1/z^a )∫_0 ^∞ sin(u)u^(aβˆ’1) du]  from euler formula  sin(x)=Im(e^(βˆ’ix) )  (d/da)∣_(a=1) [(1/z^a )Im∫_0 ^∞ e^(βˆ’iu) u^(aβˆ’1) du]  let ui=k du=(dk/i)  (d/da)∣_(a=1) [(1/z^a )Im∫_0 ^∞ e^(βˆ’k) ((k/i))^(aβˆ’1) (dk/i)]  (d/da)∣_(a=1) [(1/z^a )Im((1/i))πšͺ(a)]=(d/da)∣_(a=1) [((βˆ’πšͺ(a)sin(((𝛑a)/2)))/z^a )]  [βˆ’(z^(βˆ’a) /2)πšͺ(a)(2𝛑sin(((𝛑a)/2))(log(z)βˆ’π›™(a))βˆ’π›‘cos(((𝛑a)/2)))]_(a=1)   βˆ’(z^(βˆ’1) /2)[2𝛑(log(z)+𝛄)=(𝛑/z)log(z)+(𝛄/z)𝛑   here z=3  (βˆ’(1/4))((𝛑/3)log(3)+(𝛄/3)𝛑)βˆ’(3/8)π𝛄=βˆ’(𝛑/(12))log(3)βˆ’((𝛄𝛑)/(12))βˆ’((3𝛑)/8)Ξ³=(𝛑/4)(((log(3))/3)βˆ’(𝛄/3)βˆ’((3𝛄)/4))=(𝛑/4)(((log(3))/3)βˆ’((7𝛄)/(12)))=(𝛑/(12))(log(3)βˆ’((7𝛄)/4))
$${we}\:{know}\:{that}\:{sin}^{\mathrm{3}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{sin}\left({x}\right)βˆ’{sin}\left(\mathrm{3}{x}\right)\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}βˆ’\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{βˆ’\boldsymbol{\gamma\pi}}{\mathrm{2}}\right)βˆ’\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$$$βˆ’\frac{\mathrm{3}}{\mathrm{8}}\boldsymbol{\gamma\pi}βˆ’\frac{\mathrm{1}}{\mathrm{4}}\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{zx}}\right)\boldsymbol{{x}}^{\boldsymbol{{a}}} }{\boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{zx}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}βˆ’\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{zx}}=\boldsymbol{\mathrm{u}}\:\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{z}}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \frac{\mathrm{1}}{\boldsymbol{\mathrm{z}}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{u}}\right)\left(\frac{\boldsymbol{\mathrm{u}}}{\boldsymbol{\mathrm{z}}}\right)^{\boldsymbol{\mathrm{a}}βˆ’\mathrm{1}} \boldsymbol{\mathrm{du}}=\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{\mathrm{1}}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\int_{\mathrm{0}} ^{\infty} \boldsymbol{{sin}}\left(\boldsymbol{{u}}\right)\boldsymbol{{u}}^{\boldsymbol{{a}}βˆ’\mathrm{1}} \boldsymbol{{du}}\right] \\ $$$$\boldsymbol{{from}}\:\boldsymbol{{euler}}\:\boldsymbol{{formula}} \\ $$$$\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{Im}}\left(\boldsymbol{{e}}^{βˆ’\boldsymbol{{ix}}} \right) \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{\mathrm{1}}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\boldsymbol{{Im}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{βˆ’\boldsymbol{{iu}}} \boldsymbol{{u}}^{\boldsymbol{{a}}βˆ’\mathrm{1}} \boldsymbol{{du}}\right] \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{ui}}=\boldsymbol{{k}}\:\boldsymbol{{du}}=\frac{\boldsymbol{{dk}}}{\boldsymbol{{i}}} \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{\mathrm{1}}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\boldsymbol{{Im}}\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{βˆ’\boldsymbol{{k}}} \left(\frac{\boldsymbol{{k}}}{\boldsymbol{{i}}}\right)^{\boldsymbol{{a}}βˆ’\mathrm{1}} \frac{\boldsymbol{{dk}}}{\boldsymbol{{i}}}\right] \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{\mathrm{1}}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\boldsymbol{{Im}}\left(\frac{\mathrm{1}}{\boldsymbol{{i}}}\right)\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\right]=\frac{\boldsymbol{{d}}}{\boldsymbol{{da}}}\mid_{\boldsymbol{{a}}=\mathrm{1}} \left[\frac{βˆ’\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)}{\boldsymbol{{z}}^{\boldsymbol{{a}}} }\right] \\ $$$$\left[βˆ’\frac{{z}^{βˆ’{a}} }{\mathrm{2}}\boldsymbol{\Gamma}\left(\boldsymbol{{a}}\right)\left(\mathrm{2}\boldsymbol{\pi{sin}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)\left(\boldsymbol{{log}}\left(\boldsymbol{{z}}\right)βˆ’\boldsymbol{\psi}\left(\boldsymbol{{a}}\right)\right)βˆ’\boldsymbol{\pi{cos}}\left(\frac{\boldsymbol{\pi{a}}}{\mathrm{2}}\right)\right)\right]_{\boldsymbol{{a}}=\mathrm{1}} \\ $$$$βˆ’\frac{\boldsymbol{{z}}^{βˆ’\mathrm{1}} }{\mathrm{2}}\left[\mathrm{2}\boldsymbol{\pi}\left(\boldsymbol{{log}}\left(\boldsymbol{{z}}\right)+\boldsymbol{\gamma}\right)=\frac{\boldsymbol{\pi}}{\boldsymbol{{z}}}\boldsymbol{{log}}\left(\boldsymbol{{z}}\right)+\frac{\boldsymbol{\gamma}}{\boldsymbol{{z}}}\boldsymbol{\pi}\:\right. \\ $$$$\boldsymbol{{here}}\:\boldsymbol{{z}}=\mathrm{3} \\ $$$$\left(βˆ’\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\boldsymbol{\pi}}{\mathrm{3}}\boldsymbol{{log}}\left(\mathrm{3}\right)+\frac{\boldsymbol{\gamma}}{\mathrm{3}}\boldsymbol{\pi}\right)βˆ’\frac{\mathrm{3}}{\mathrm{8}}\pi\boldsymbol{\gamma}=βˆ’\frac{\boldsymbol{\pi}}{\mathrm{12}}\boldsymbol{{log}}\left(\mathrm{3}\right)βˆ’\frac{\boldsymbol{\gamma\pi}}{\mathrm{12}}βˆ’\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{8}}\gamma=\frac{\boldsymbol{\pi}}{\mathrm{4}}\left(\frac{\boldsymbol{{log}}\left(\mathrm{3}\right)}{\mathrm{3}}βˆ’\frac{\boldsymbol{\gamma}}{\mathrm{3}}βˆ’\frac{\mathrm{3}\boldsymbol{\gamma}}{\mathrm{4}}\right)=\frac{\boldsymbol{\pi}}{\mathrm{4}}\left(\frac{\boldsymbol{{log}}\left(\mathrm{3}\right)}{\mathrm{3}}βˆ’\frac{\mathrm{7}\boldsymbol{\gamma}}{\mathrm{12}}\right)=\frac{\boldsymbol{\pi}}{\mathrm{12}}\left(\boldsymbol{{log}}\left(\mathrm{3}\right)βˆ’\frac{\mathrm{7}\boldsymbol{\gamma}}{\mathrm{4}}\right) \\ $$$$ \\ $$

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