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Question Number 132987 by mnjuly1970 last updated on 17/Feb/21

....mathematical analysis... prove that:: 𝛗=∫_0 ^( ∞) ((sin^3 (x))/x^3 )dx=((3Ο€)/8) βˆ—βˆ—βˆ—βˆ—..........

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{mathematical}\:\:{analysis}... \\ $$$$\:{prove}\:\:{that}::\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{3}} \left({x}\right)}{{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\:\:\:\:\ast\ast\ast\ast.......... \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 17/Feb/21

∫_0 ^∞ ((sin^3 x)/x^3 )dx =(3/4)∫_0 ^∞ ((sinx)/x^3 )βˆ’(1/4)∫_0 ^∞ ((sin3x)/x^3 )dx =(3/4).(Ο€/(4sin(((3Ο€)/2))))βˆ’(9/4)∫_0 ^∞ ((sinu)/u^3 )du ∫_0 ^∞ ((sinx)/x^ΞΌ )du=(Ο€/(2Ξ“(ΞΌ)sin((Ο€/2)ΞΌ))) =βˆ’(9/(4.4sin(((3Ο€)/2))))βˆ’((3Ο€)/(16))=((9Ο€)/(16))βˆ’((3Ο€)/(16))=((3Ο€)/8)

$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{3}} {x}}{{x}^{\mathrm{3}} }{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\mathrm{3}} }βˆ’\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\mathrm{3}{x}}{{x}^{\mathrm{3}} }{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}.\frac{\pi}{\mathrm{4}{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)}βˆ’\frac{\mathrm{9}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sinu}}{{u}^{\mathrm{3}} }{du}\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\mu} }{du}=\frac{\pi}{\mathrm{2}\Gamma\left(\mu\right){sin}\left(\frac{\pi}{\mathrm{2}}\mu\right)} \\ $$$$=βˆ’\frac{\mathrm{9}}{\mathrm{4}.\mathrm{4}{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)}βˆ’\frac{\mathrm{3}\pi}{\mathrm{16}}=\frac{\mathrm{9}\pi}{\mathrm{16}}βˆ’\frac{\mathrm{3}\pi}{\mathrm{16}}=\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$

Commented by mnjuly1970 last updated on 18/Feb/21

thanks alot mr power...

$${thanks}\:{alot}\:{mr}\:{power}... \\ $$

Commented by Dwaipayan Shikari last updated on 18/Feb/21

Mr power?? hahaaa....sir

$${Mr}\:{power}?? \\ $$$${hahaaa}....{sir} \\ $$

Commented by mnjuly1970 last updated on 18/Feb/21

sorry sorry... master Dwaipayan..

$$\:{sorry}\:\:{sorry}... \\ $$$$\:\:{master}\:{Dwaipayan}.. \\ $$$$\:\: \\ $$

Answered by mnjuly1970 last updated on 18/Feb/21

another solution... Ο•(a)=∫_0 ^( ∞) ((sin^3 (ax))/x^3 )dx 𝛗=Ο•(1) ... Ο•β€²(a)=(βˆ‚/βˆ‚a)(∫_0 ^( ∞) ((sin^3 (ax))/x^3 )dx) =∫_0 ^( ∞) ((3xcos(ax)sin^2 (ax))/x^3 )dx =3∫_0 ^( ∞) ((sin^2 (ax)cos(ax))/x^2 ) dx =(3/2)∫_0 ^( ∞) ((sin(2ax)sin(ax))/x^2 )dx =(3/4)∫_0 ^( ∞) ((cos(ax)βˆ’cos(3ax))/x^2 )dx Ο•β€²β€²(a)=(3/4)∫_0 ^( ∞) ((βˆ’sin(ax))/x)dx+(9/4)∫_0 ^( ∞) ((sin(3ax))/x)dx =((3Ο€)/4) β‡’ Ο•β€²(a)=(3/4)Ο€a+C Ο•β€²(0)=0=C Ο•β€²(a)=(3/4)Ο€a β‡’ Ο•(a)=(3/8)Ο€a^2 +K Ο•(0)=0=K Ο•(a)=(3/8)Ο€a^2 ⇒𝛗=Ο•(1)=(3/8)Ο€ βœ“βœ“

$$\:\:\:{another}\:\:{solution}... \\ $$$$\:\:\:\varphi\left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{3}} \left({ax}\right)}{{x}^{\mathrm{3}} }{dx} \\ $$$$\:\:\:\:\boldsymbol{\phi}=\varphi\left(\mathrm{1}\right)\:... \\ $$$$\:\:\:\:\varphi'\left({a}\right)=\frac{\partial}{\partial{a}}\left(\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{3}} \left({ax}\right)}{{x}^{\mathrm{3}} }{dx}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{3}{xcos}\left({ax}\right){sin}^{\mathrm{2}} \left({ax}\right)}{{x}^{\mathrm{3}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{3}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{2}} \left({ax}\right){cos}\left({ax}\right)}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{2}{ax}\right){sin}\left({ax}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({ax}\right)βˆ’{cos}\left(\mathrm{3}{ax}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\varphi''\left({a}\right)=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{βˆ’{sin}\left({ax}\right)}{{x}}{dx}+\frac{\mathrm{9}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{3}{ax}\right)}{{x}}{dx} \\ $$$$=\frac{\mathrm{3}\pi}{\mathrm{4}}\:\:\:\Rightarrow\:\varphi'\left({a}\right)=\frac{\mathrm{3}}{\mathrm{4}}\pi{a}+{C} \\ $$$$\:\:\:\:\varphi'\left(\mathrm{0}\right)=\mathrm{0}={C} \\ $$$$\:\:\:\varphi'\left({a}\right)=\frac{\mathrm{3}}{\mathrm{4}}\pi{a}\:\Rightarrow\:\varphi\left({a}\right)=\frac{\mathrm{3}}{\mathrm{8}}\pi{a}^{\mathrm{2}} +{K} \\ $$$$\:\:\:\:\varphi\left(\mathrm{0}\right)=\mathrm{0}={K} \\ $$$$\:\:\:\:\:\:\varphi\left({a}\right)=\frac{\mathrm{3}}{\mathrm{8}}\pi{a}^{\mathrm{2}} \:\:\Rightarrow\boldsymbol{\phi}=\varphi\left(\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{8}}\pi\:\checkmark\checkmark \\ $$

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