Question Number 121674 by mnjuly1970 last updated on 10/Nov/20
$$\:\:\:\:\:\:\:\:\:…\:\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:{prove}\:{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\Omega\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{4}} \left({x}\right){ln}\left({x}\right)}{{x}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−\gamma\right) \\ $$$$\gamma:{euler}−{mascheroni}\:{constant} \\ $$$$\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970} \\ $$
Answered by mindispower last updated on 11/Nov/20
![Ω=∫_0 ^∞ ((sin^4 (x))/x^2 )ln(x)dx sin^4 (x)=(1/8)(cos(4x)−4cos(2x)+3) 8Ω=∫_0 ^∞ (cos(4x)−4cos(2x)+3).((ln(x)dx)/x^2 ) by part u=cos(4x)−4cos(2x)+3 dv=((ln(x))/x^2 ),v=−((ln(x))/x)−(1/x) 8Ω=[−(((ln(x)+1))/x)(cos(4x)−4cos(2x)+3)]_0 ^∞ +∫_0 ^∞ (((ln(x)+1))/x)(−4sin(4x)+8sin(2x))dx 8Ω=−4∫_0 ^∞ ((ln(x)+1)/x)(sin(4x)−2sin(2x))dx ⇔−2Ω={∫_0 ^∞ ((ln(x))/x)sin(4x)−2∫((ln(x))/x)sin(2x)}_(=W) +{∫_0 ^∞ ((sin(4x))/x)−2∫_0 ^∞ ((sin(2x))/x)dx}_(=V) a>0,∫_0 ^∞ ((sin(ax))/x)dx=∫_0 ^∞ ((sin(ax))/(ax))d(ax)=(π/2) V=(π/2)−2.(π/2)=−(π/2) ∫_0 ^∞ ((sin(ax))/x)ln(x)=f(a)=(∂/∂t)∫_0 ^∞ sin(ax)x^(t−1) dx∣_(t=0) f(t)=∫_0 ^∞ sin(ax)x^(t−1) dx=Im∫_0 ^∞ e^(iax) x^(t−1) dx iax=−z⇔Im∫_0 ^(i∞) e^(−z) .(((iz)/a))^t .(a/z).(dz/a) =Ima∫_0 ^(i∞) e^(i(π/2)t) ((z/a))^(t−1) e^(−z) dz =Im−(e^(i(π/2)t) /a^t )∫_0 ^∞ z^(t−1) e^(−z) dz =−((sin(((πt)/2))Γ(t))/a^t )=((−sin(((πt)/2))π)/(a^t Γ(1−t)sin(((πt)/2))))=−(π/(2cos((π/2)t)Γ(1−t)a^t )) f′(t)∣_(t=0) =(π/2)(−γ−ln(a)))=−(π/2)(γ+ln(a)) W=f(4)−2f(2)=−(π/2)(γ+ln(4))−2(−(π/2)(γ+ln(2)) =(π/2)γ −2Ω=W+V=(π/2)γ−(π/2)⇒Ω=−((πγ)/4)+(π/4)=(π/4)(1−γ)](https://www.tinkutara.com/question/Q121717.png)
$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{4}} \left({x}\right)}{{x}^{\mathrm{2}} }{ln}\left({x}\right){dx} \\ $$$${sin}^{\mathrm{4}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left({cos}\left(\mathrm{4}{x}\right)−\mathrm{4}{cos}\left(\mathrm{2}{x}\right)+\mathrm{3}\right) \\ $$$$\mathrm{8}\Omega=\int_{\mathrm{0}} ^{\infty} \left({cos}\left(\mathrm{4}{x}\right)−\mathrm{4}{cos}\left(\mathrm{2}{x}\right)+\mathrm{3}\right).\frac{{ln}\left({x}\right){dx}}{{x}^{\mathrm{2}} } \\ $$$${by}\:{part}\:{u}={cos}\left(\mathrm{4}{x}\right)−\mathrm{4}{cos}\left(\mathrm{2}{x}\right)+\mathrm{3} \\ $$$${dv}=\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} },{v}=−\frac{{ln}\left({x}\right)}{{x}}−\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{8}\Omega=\left[−\frac{\left({ln}\left({x}\right)+\mathrm{1}\right)}{{x}}\left({cos}\left(\mathrm{4}{x}\right)−\mathrm{4}{cos}\left(\mathrm{2}{x}\right)+\mathrm{3}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$+\int_{\mathrm{0}} ^{\infty} \frac{\left({ln}\left({x}\right)+\mathrm{1}\right)}{{x}}\left(−\mathrm{4}{sin}\left(\mathrm{4}{x}\right)+\mathrm{8}{sin}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$\mathrm{8}\Omega=−\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)+\mathrm{1}}{{x}}\left({sin}\left(\mathrm{4}{x}\right)−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$\Leftrightarrow−\mathrm{2}\Omega=\left\{\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{{x}}{sin}\left(\mathrm{4}{x}\right)−\mathrm{2}\int\frac{{ln}\left({x}\right)}{{x}}{sin}\left(\mathrm{2}{x}\right)\right\}_{={W}} \\ $$$$+\left\{\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\mathrm{4}{x}\right)}{{x}}−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx}\right\}_{={V}} \\ $$$${a}>\mathrm{0},\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({ax}\right)}{{x}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({ax}\right)}{{ax}}{d}\left({ax}\right)=\frac{\pi}{\mathrm{2}} \\ $$$${V}=\frac{\pi}{\mathrm{2}}−\mathrm{2}.\frac{\pi}{\mathrm{2}}=−\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({ax}\right)}{{x}}{ln}\left({x}\right)={f}\left({a}\right)=\frac{\partial}{\partial{t}}\int_{\mathrm{0}} ^{\infty} {sin}\left({ax}\right){x}^{{t}−\mathrm{1}} {dx}\mid_{{t}=\mathrm{0}} \\ $$$${f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} {sin}\left({ax}\right){x}^{{t}−\mathrm{1}} {dx}={Im}\int_{\mathrm{0}} ^{\infty} {e}^{{iax}} {x}^{{t}−\mathrm{1}} {dx} \\ $$$${iax}=−{z}\Leftrightarrow{Im}\int_{\mathrm{0}} ^{{i}\infty} {e}^{−{z}} .\left(\frac{{iz}}{{a}}\right)^{{t}} .\frac{{a}}{{z}}.\frac{{dz}}{{a}} \\ $$$$={Ima}\int_{\mathrm{0}} ^{{i}\infty} {e}^{{i}\frac{\pi}{\mathrm{2}}{t}} \left(\frac{{z}}{{a}}\right)^{{t}−\mathrm{1}} {e}^{−{z}} {dz} \\ $$$$={Im}−\frac{{e}^{{i}\frac{\pi}{\mathrm{2}}{t}} }{{a}^{{t}} }\int_{\mathrm{0}} ^{\infty} {z}^{{t}−\mathrm{1}} {e}^{−{z}} {dz} \\ $$$$=−\frac{{sin}\left(\frac{\pi{t}}{\mathrm{2}}\right)\Gamma\left({t}\right)}{{a}^{{t}} }=\frac{−{sin}\left(\frac{\pi{t}}{\mathrm{2}}\right)\pi}{{a}^{{t}} \Gamma\left(\mathrm{1}−{t}\right){sin}\left(\frac{\pi{t}}{\mathrm{2}}\right)}=−\frac{\pi}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}}{t}\right)\Gamma\left(\mathrm{1}−{t}\right){a}^{{t}} } \\ $$$$\left.{f}'\left({t}\right)\mid_{{t}=\mathrm{0}} =\frac{\pi}{\mathrm{2}}\left(−\gamma−{ln}\left({a}\right)\right)\right)=−\frac{\pi}{\mathrm{2}}\left(\gamma+{ln}\left({a}\right)\right) \\ $$$${W}={f}\left(\mathrm{4}\right)−\mathrm{2}{f}\left(\mathrm{2}\right)=−\frac{\pi}{\mathrm{2}}\left(\gamma+{ln}\left(\mathrm{4}\right)\right)−\mathrm{2}\left(−\frac{\pi}{\mathrm{2}}\left(\gamma+{ln}\left(\mathrm{2}\right)\right)\right. \\ $$$$=\frac{\pi}{\mathrm{2}}\gamma \\ $$$$−\mathrm{2}\Omega={W}+{V}=\frac{\pi}{\mathrm{2}}\gamma−\frac{\pi}{\mathrm{2}}\Rightarrow\Omega=−\frac{\pi\gamma}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−\gamma\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 11/Nov/20
$${peace}\:{be}\:{upon}\:{you} \\ $$$${mr}\:{power}\:.{thank}\:{you}. \\ $$
Commented by mindispower last updated on 11/Nov/20
$${withe}\:{pleasur}\:{sir}\: \\ $$
Answered by mnjuly1970 last updated on 11/Nov/20
Answered by mnjuly1970 last updated on 11/Nov/20
