... advanced calculus... prove that : Ω =∫_0 ^( ∞) ((sin^4 (x)ln(x))/x^2 )dx=(π/4)(1−γ) γ:euler−mascheroni constant m.n.july.1970


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Question Number 121674 by mnjuly1970 last updated on 10/Nov/20

$. . . a d v a n c e d c a l c u l u s . . .$ $p r o v e t h a t :$ $Ω = \int_{0}^{\infty} \frac{{s i n}^{4} (x) l n (x)}{x^{2}} d x = \frac{π}{4} (1 - γ)$ $γ : e u l e r - m a s c h e r o n i c o n s t a n t$ $m . n . j u l y . 1970$
	Answered by mindispower last updated on 11/Nov/20
	$Ω=∫_0 ^∞ ((sin^4 (x))/x^2 )ln(x)dx sin^4 (x)=(1/8)(cos(4x)−4cos(2x)+3) 8Ω=∫_0 ^∞ (cos(4x)−4cos(2x)+3).((ln(x)dx)/x^2 ) by part u=cos(4x)−4cos(2x)+3 dv=((ln(x))/x^2 ),v=−((ln(x))/x)−(1/x) 8Ω=[−(((ln(x)+1))/x)(cos(4x)−4cos(2x)+3)]_0 ^∞ +∫_0 ^∞ (((ln(x)+1))/x)(−4sin(4x)+8sin(2x))dx 8Ω=−4∫_0 ^∞ ((ln(x)+1)/x)(sin(4x)−2sin(2x))dx ⇔−2Ω={∫_0 ^∞ ((ln(x))/x)sin(4x)−2∫((ln(x))/x)sin(2x)}_(=W) +{∫_0 ^∞ ((sin(4x))/x)−2∫_0 ^∞ ((sin(2x))/x)dx}_(=V) a>0,∫_0 ^∞ ((sin(ax))/x)dx=∫_0 ^∞ ((sin(ax))/(ax))d(ax)=(π/2) V=(π/2)−2.(π/2)=−(π/2) ∫_0 ^∞ ((sin(ax))/x)ln(x)=f(a)=(∂/∂t)∫_0 ^∞ sin(ax)x^(t−1) dx∣_(t=0) f(t)=∫_0 ^∞ sin(ax)x^(t−1) dx=Im∫_0 ^∞ e^(iax) x^(t−1) dx iax=−z⇔Im∫_0 ^(i∞) e^(−z) .(((iz)/a))^t .(a/z).(dz/a) =Ima∫_0 ^(i∞) e^(i(π/2)t) ((z/a))^(t−1) e^(−z) dz =Im−(e^(i(π/2)t) /a^t )∫_0 ^∞ z^(t−1) e^(−z) dz =−((sin(((πt)/2))Γ(t))/a^t )=((−sin(((πt)/2))π)/(a^t Γ(1−t)sin(((πt)/2))))=−(π/(2cos((π/2)t)Γ(1−t)a^t )) f′(t)∣_(t=0) =(π/2)(−γ−ln(a)))=−(π/2)(γ+ln(a)) W=f(4)−2f(2)=−(π/2)(γ+ln(4))−2(−(π/2)(γ+ln(2)) =(π/2)γ −2Ω=W+V=(π/2)γ−(π/2)⇒Ω=−((πγ)/4)+(π/4)=(π/4)(1−γ)$
	$Ω = \int_{0}^{\infty} \frac{{s i n}^{4} (x)}{x^{2}} l n (x) d x$ ${s i n}^{4} (x) = \frac{1}{8} (c o s (4 x) - 4 c o s (2 x) + 3)$ $8 Ω = \int_{0}^{\infty} (c o s (4 x) - 4 c o s (2 x) + 3) . \frac{l n (x) d x}{x^{2}}$ $b y p a r t u = c o s (4 x) - 4 c o s (2 x) + 3$ $d v = \frac{l n (x)}{x^{2}}, v = - \frac{l n (x)}{x} - \frac{1}{x}$ $8 Ω = {[- \frac{(l n (x) + 1)}{x} (c o s (4 x) - 4 c o s (2 x) + 3)]}_{0}^{\infty}$ $+ \int_{0}^{\infty} \frac{(l n (x) + 1)}{x} (- 4 s i n (4 x) + 8 s i n (2 x)) d x$ $8 Ω = - 4 \int_{0}^{\infty} \frac{l n (x) + 1}{x} (s i n (4 x) - 2 s i n (2 x)) d x$ $\Leftrightarrow - 2 Ω = {\int_{0}^{\infty} \frac{l n (x)}{x} s i n (4 x) - 2 \int \frac{l n (x)}{x} s i n (2 x)}_{= W}$ $+ {\int_{0}^{\infty} \frac{s i n (4 x)}{x} - 2 \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x}_{= V}$ $a > 0, \int_{0}^{\infty} \frac{s i n (a x)}{x} d x = \int_{0}^{\infty} \frac{s i n (a x)}{a x} d (a x) = \frac{π}{2}$ $V = \frac{π}{2} - 2 . \frac{π}{2} = - \frac{π}{2}$ $\int_{0}^{\infty} \frac{s i n (a x)}{x} l n (x) = f (a) = \frac{\partial}{\partial t} \int_{0}^{\infty} s i n (a x) x^{t - 1} d x ∣_{t = 0}$ $f (t) = \int_{0}^{\infty} s i n (a x) x^{t - 1} d x = I m \int_{0}^{\infty} e^{i a x} x^{t - 1} d x$ $i a x = - z \Leftrightarrow I m \int_{0}^{i \infty} e^{- z} . {(\frac{i z}{a})}^{t} . \frac{a}{z} . \frac{d z}{a}$ $= I m a \int_{0}^{i \infty} e^{i \frac{π}{2} t} {(\frac{z}{a})}^{t - 1} e^{- z} d z$ $= I m - \frac{e^{i \frac{π}{2} t}}{a^{t}} \int_{0}^{\infty} z^{t - 1} e^{- z} d z$ $= - \frac{s i n (\frac{π t}{2}) Γ (t)}{a^{t}} = \frac{- s i n (\frac{π t}{2}) π}{a^{t} Γ (1 - t) s i n (\frac{π t}{2})} = - \frac{π}{2 c o s (\frac{π}{2} t) Γ (1 - t) a^{t}}$ $f^{'} (t) ∣_{t = 0} = \frac{π}{2} (- γ - l n (a))) = - \frac{π}{2} (γ + l n (a))$ $W = f (4) - 2 f (2) = - \frac{π}{2} (γ + l n (4)) - 2 (- \frac{π}{2} (γ + l n (2))$ $= \frac{π}{2} γ$ $- 2 Ω = W + V = \frac{π}{2} γ - \frac{π}{2} \Rightarrow Ω = - \frac{π γ}{4} + \frac{π}{4} = \frac{π}{4} (1 - γ)$
	Commented by mnjuly1970 last updated on 11/Nov/20

	$p e a c e b e u p o n y o u$ $m r p o w e r . t h a n k y o u .$
	Commented by mindispower last updated on 11/Nov/20

	$w i t h e p l e a s u r s i r$
	Answered by mnjuly1970 last updated on 11/Nov/20

	Answered by mnjuly1970 last updated on 11/Nov/20

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