Question Number 140956 by mnjuly1970 last updated on 14/May/21
$$\:\:\:\:\:\:\:\:\:\:.....{advanced}......{calculus}..... \\ $$$$\:\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\boldsymbol{\phi}:=\:\int_{−\infty} ^{\:\infty} \frac{{sin}^{\mathrm{4}} \left({x}\right).{cos}^{\mathrm{4}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{16}} \\ $$$$\:\:{m}.{n} \\ $$
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))★](Q140962.png)
$$\Phi\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{sinx}.\mathrm{cosx}\right)^{\mathrm{4}} }{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{4}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}^{\mathrm{4}} \left(\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \:\frac{\left(\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{32}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−\mathrm{2cos}\left(\mathrm{4x}\right)+\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{8x}\right)}{\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}−\mathrm{4cos}\left(\mathrm{4x}\right)+\mathrm{1}+\mathrm{cos}\left(\mathrm{8x}\right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{4}−\mathrm{4cos}\left(\mathrm{4x}\right)−\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{8x}\right)\right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{8x}\right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\mathrm{we}\:\mathrm{have}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=_{\mathrm{2x}=\mathrm{t}} \:\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\mathrm{cos2t}}{\mathrm{t}^{\mathrm{2}} }\frac{\mathrm{dt}}{\mathrm{2}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{t}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\left\{\:\left[−\frac{\mathrm{1}}{\mathrm{t}}\mathrm{sin}^{\mathrm{2}} \mathrm{t}\right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2sint}.\mathrm{cost}}{\mathrm{t}}\:\mathrm{dt}\right\} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{2t}\right)}{\mathrm{t}}\:\mathrm{dt}\:\:=_{\mathrm{2t}=\mathrm{z}} \:\:\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinz}}{\frac{\mathrm{z}}{\mathrm{2}}}\frac{\mathrm{dz}}{\mathrm{2}}\:=\mathrm{4}.\frac{\pi}{\mathrm{2}}=\mathrm{2}\pi \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{8x}\right)}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=_{\mathrm{4x}=\mathrm{t}} \mathrm{16}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2t}\right)}{\mathrm{t}^{\mathrm{2}} }\frac{\mathrm{dt}}{\mathrm{4}} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{t}^{\mathrm{2}} }\:=\mathrm{8}.\frac{\pi}{\mathrm{2}}=\mathrm{4}\pi\:\Rightarrow \\ $$$$\Phi\:=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2}\pi\right)−\frac{\mathrm{1}}{\mathrm{64}}\left(\mathrm{4}\pi\right)\:=\frac{\pi}{\mathrm{8}}−\frac{\pi}{\mathrm{16}}\:=\frac{\pi}{\mathrm{16}}\:\Rightarrow\Phi=\frac{\pi}{\mathrm{16}}\bigstar \\ $$
Commented by mnjuly1970 last updated on 14/May/21
$$\:\:\:\:{grateful}\:{sir}\:{max}\:...{thanking} \\ $$