.....advanced......calculus..... prove that: 𝛗:= ∫_(−∞) ^( ∞) ((sin^4 (x).cos^4 (x))/x^2 )dx=(π/(16)) m.n


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 140956 by mnjuly1970 last updated on 14/May/21

$. . . . . a d v a n c e d . . . . . . c a l c u l u s . . . . .$ $p r o v e t h a t :$ $ϕ := \int_{- \infty}^{\infty} \frac{{s i n}^{4} (x) . {c o s}^{4} (x)}{x^{2}} d x = \frac{π}{16}$ $m . n$
	Answered by mathmax by abdo last updated on 14/May/21
	$Φ =2∫_0 ^∞ (((sinx.cosx)^4 )/x^2 )dx =(2/2^4 )∫_0 ^∞ ((sin^4 (2x))/x^2 )dx =(1/8)∫_0 ^∞ (((((1−cos(4x))/2))^2 )/x^2 )dx =(1/(32))∫_0 ^∞ ((1−2cos(4x)+((1+cos(8x))/2))/x^2 )dx =(1/(64))∫_0 ^∞ ((2−4cos(4x)+1+cos(8x))/x^2 )dx =(1/(64))∫_0 ^∞ ((4−4cos(4x)−(1−cos(8x)))/x^2 )dx =(1/(16))∫_0 ^∞ ((1−cos(4x))/x^2 )dx−(1/(64))∫_0 ^∞ ((1−cos(8x))/x^2 )dx we have ∫_0 ^∞ ((1−cos(4x))/x^2 )dx =_(2x=t) 4 ∫_0 ^∞ ((1−cos2t)/t^2 )(dt/2) =2∫_0 ^∞ ((2sin^2 (t))/t^2 )dt =4∫_0 ^∞ ((sin^2 (t))/t^2 ) =4{ [−(1/t)sin^2 t]_0 ^∞ +∫_0 ^∞ ((2sint.cost)/t) dt} =4∫_0 ^∞ ((sin(2t))/t) dt =_(2t=z) 4∫_0 ^∞ ((sinz)/(z/2))(dz/2) =4.(π/2)=2π ∫_0 ^∞ ((1−cos(8x))/x^2 ) dx =_(4x=t) 16∫_0 ^∞ ((1−cos(2t))/t^2 )(dt/4) =4 ∫_0 ^∞ ((2sin^2 t)/t^2 ) dt =8 ∫_0 ^∞ ((sin^2 t)/t^2 ) =8.(π/2)=4π ⇒ Φ =(1/(16))(2π)−(1/(64))(4π) =(π/8)−(π/(16)) =(π/(16)) ⇒Φ=(π/(16))★$
	$Φ = 2 \int_{0}^{\infty} \frac{{(sinx . cosx)}^{4}}{x^{2}} dx = \frac{2}{2^{4}} \int_{0}^{\infty} \frac{\sin^{4} (2 x)}{x^{2}} dx$ $= \frac{1}{8} \int_{0}^{\infty} \frac{{(\frac{1 - \cos (4 x)}{2})}^{2}}{x^{2}} dx = \frac{1}{32} \int_{0}^{\infty} \frac{1 - 2 \cos (4 x) + \frac{1 + \cos (8 x)}{2}}{x^{2}} dx$ $= \frac{1}{64} \int_{0}^{\infty} \frac{2 - 4 \cos (4 x) + 1 + \cos (8 x)}{x^{2}} dx$ $= \frac{1}{64} \int_{0}^{\infty} \frac{4 - 4 \cos (4 x) - (1 - \cos (8 x))}{x^{2}} dx$ $= \frac{1}{16} \int_{0}^{\infty} \frac{1 - \cos (4 x)}{x^{2}} dx - \frac{1}{64} \int_{0}^{\infty} \frac{1 - \cos (8 x)}{x^{2}} dx$ $we have \int_{0}^{\infty} \frac{1 - \cos (4 x)}{x^{2}} dx =_{2 x = t} 4 \int_{0}^{\infty} \frac{1 - \cos 2 t}{t^{2}} \frac{dt}{2}$ $= 2 \int_{0}^{\infty} \frac{{2 \sin}^{2} (t)}{t^{2}} dt = 4 \int_{0}^{\infty} \frac{\sin^{2} (t)}{t^{2}}$ $= 4 {{[- \frac{1}{t} \sin^{2} t]}_{0}^{\infty} + \int_{0}^{\infty} \frac{2 sint . cost}{t} dt}$ $= 4 \int_{0}^{\infty} \frac{\sin (2 t)}{t} dt =_{2 t = z} 4 \int_{0}^{\infty} \frac{sinz}{\frac{z}{2}} \frac{dz}{2} = 4 . \frac{π}{2} = 2 π$ $\int_{0}^{\infty} \frac{1 - \cos (8 x)}{x^{2}} dx =_{4 x = t} 16 \int_{0}^{\infty} \frac{1 - \cos (2 t)}{t^{2}} \frac{dt}{4}$ $= 4 \int_{0}^{\infty} \frac{{2 \sin}^{2} t}{t^{2}} dt = 8 \int_{0}^{\infty} \frac{\sin^{2} t}{t^{2}} = 8 . \frac{π}{2} = 4 π \Rightarrow$ $Φ = \frac{1}{16} (2 π) - \frac{1}{64} (4 π) = \frac{π}{8} - \frac{π}{16} = \frac{π}{16} \Rightarrow Φ = \frac{π}{16} ★$
	Commented by mnjuly1970 last updated on 14/May/21

	$g r a t e f u l s i r m a x . . . t h a n k i n g$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum