Question Number 143312 by Ar Brandon last updated on 12/Jun/21
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{4}} \mathrm{x}}{\mathrm{x}^{\mathrm{4}} }\mathrm{dx}=\frac{\pi}{\mathrm{3}} \\ $$
Answered by Olaf_Thorendsen last updated on 12/Jun/21
![Let f(x) = 1 (constant function unity) As f is a continuous function satisfying the π−periodic assumption, we can apply the extended Lobachevsky′s Dirichlet integral formula : ∫_0 ^∞ ((sin^4 )/x^4 )f(x) dx = ∫_0 ^(π/2) f(x) dx−(2/3)∫_0 ^(π/2) sin^2 x.f(x)dx ⇒ ∫_0 ^∞ ((sin^4 )/x^4 ) dx = ∫_0 ^(π/2) dx−(2/3)∫_0 ^(π/2) sin^2 x dx ∫_0 ^∞ ((sin^4 )/x^4 ) dx = (π/2)−(2/3)∫_0 ^(π/2) ((1−cos2x)/2) dx ∫_0 ^∞ ((sin^4 )/x^4 ) dx = (π/2)−(1/3)[x−(1/2)sin2x]_0 ^(π/2) ∫_0 ^∞ ((sin^4 )/x^4 ) dx = (π/2)−(1/3)((π/2)) ∫_0 ^∞ ((sin^4 )/x^4 ) dx = (π/2)−(π/6) = (π/3)](Q143313.png)
$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\mathrm{1}\:\left(\mathrm{constant}\:\mathrm{function}\:\mathrm{unity}\right) \\ $$$$\mathrm{As}\:{f}\:\mathrm{is}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{satisfying} \\ $$$$\mathrm{the}\:\pi−\mathrm{periodic}\:\mathrm{assumption},\:\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{apply}\:\mathrm{the}\:\mathrm{extended}\:\mathrm{Lobachevsky}'\mathrm{s} \\ $$$$\:\mathrm{Dirichlet}\:\mathrm{integral}\:\mathrm{formula}\:: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{4}} }{{x}^{\mathrm{4}} }{f}\left({x}\right)\:{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right)\:{dx}−\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} {x}.{f}\left({x}\right){dx} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{4}} }{{x}^{\mathrm{4}} }\:{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}−\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} {x}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{4}} }{{x}^{\mathrm{4}} }\:{dx}\:=\:\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−\mathrm{cos2}{x}}{\mathrm{2}}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{4}} }{{x}^{\mathrm{4}} }\:{dx}\:=\:\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\left[{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{4}} }{{x}^{\mathrm{4}} }\:{dx}\:=\:\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{4}} }{{x}^{\mathrm{4}} }\:{dx}\:=\:\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\:=\:\frac{\pi}{\mathrm{3}} \\ $$
Commented by Ar Brandon last updated on 12/Jun/21
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$
Commented by Ar Brandon last updated on 12/Jun/21
![I tried this; Φ=∫_0 ^∞ ((sin^4 x)/x^4 )dx, sin^4 x=(((1−2cos2x+cos^2 2x))/4)=(1/4)−((cos2x)/2)+(((1+cos4x))/8) Φ=(3/8)∫_0 ^∞ (1/x^4 )dx−(1/2)∫_0 ^∞ ((cos2x)/x^4 )dx+(1/8)∫_0 ^∞ ((cos4x)/x^4 )dx =−[(1/(8x^3 ))]_0 ^∞ −((π2^3 )/(4Γ(4)cos((π/2)×4)))+((π4^3 )/(16Γ(4)cos((π/2)×4))) =−[(1/(8x^3 ))]_0 ^∞ −((8π)/(24))+((64π)/(48))=... Why did it not go ? Do you know why ?](Q143315.png)
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{this}; \\ $$$$\Phi=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{4}} \mathrm{x}}{\mathrm{x}^{\mathrm{4}} }\mathrm{dx},\: \\ $$$$\mathrm{sin}^{\mathrm{4}} \mathrm{x}=\frac{\left(\mathrm{1}−\mathrm{2cos2x}+\mathrm{cos}^{\mathrm{2}} \mathrm{2x}\right)}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{cos2x}}{\mathrm{2}}+\frac{\left(\mathrm{1}+\mathrm{cos4x}\right)}{\mathrm{8}} \\ $$$$\Phi=\frac{\mathrm{3}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos2x}}{\mathrm{x}^{\mathrm{4}} }\mathrm{dx}+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos4x}}{\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$$$\:\:\:\:=−\left[\frac{\mathrm{1}}{\mathrm{8x}^{\mathrm{3}} }\right]_{\mathrm{0}} ^{\infty} −\frac{\pi\mathrm{2}^{\mathrm{3}} }{\mathrm{4}\Gamma\left(\mathrm{4}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}×\mathrm{4}\right)}+\frac{\pi\mathrm{4}^{\mathrm{3}} }{\mathrm{16}\Gamma\left(\mathrm{4}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}×\mathrm{4}\right)} \\ $$$$\:\:\:\:\:=−\left[\frac{\mathrm{1}}{\mathrm{8x}^{\mathrm{3}} }\right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{8}\pi}{\mathrm{24}}+\frac{\mathrm{64}\pi}{\mathrm{48}}=... \\ $$$$\mathrm{Why}\:\mathrm{did}\:\mathrm{it}\:\mathrm{not}\:\mathrm{go}\:?\:\mathrm{Do}\:\mathrm{you}\:\mathrm{know}\:\mathrm{why}\:? \\ $$