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Question Number 138860 by qaz last updated on 19/Apr/21
Prove:: ∫_0 ^∞ ((x^3 −sin^3 x)/x^5 )dx=((13)/(32))π
$${Prove}::\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} −\mathrm{sin}\:^{\mathrm{3}} {x}}{{x}^{\mathrm{5}} }{dx}=\frac{\mathrm{13}}{\mathrm{32}}\pi \\ $$
Answered by Dwaipayan Shikari last updated on 19/Apr/21
∫_0 ^∞ ((x^3 −sin^3 x)/x^5 )dx ∫_0 ^∞ ((x^3 −(3sinx−sin3x)/4)/x^5 )dx J(a)+G(3)=(1/4)∫_0 ^∞ ((4x^3 −3sin(ax))/x^5 )+(1/4)∫_0 ^∞ ((sin3x)/x^5 )dx J′′(a)=(3/4)∫_0 ^∞ ((sin(ax))/x^3 )dx=((3a^2 )/4)∫_0 ^∞ ((sin(t))/( t^3 ))dt=((3a^2 )/4).(π/(2Γ(3)sin(((3π)/2)))) =−((3a^2 π)/(16)) ⇒J(a)=−((a^4 π)/(64))+Ka+C K=0 C=0 G(3)=(3^4 /4).(π/(2Γ(5)sin(((5π)/2))))=((81π)/(192))=((27π)/(64)) J′′(1)+G(3)=((−π)/(64))+((27π)/(64))=((13π)/(32))
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} −{sin}^{\mathrm{3}} {x}}{{x}^{\mathrm{5}} }{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} −\left(\mathrm{3}{sinx}−{sin}\mathrm{3}{x}\right)/\mathrm{4}}{{x}^{\mathrm{5}} }{dx} \\ $$$${J}\left({a}\right)+{G}\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{sin}\left({ax}\right)}{{x}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\mathrm{3}{x}}{{x}^{\mathrm{5}} }{dx} \\ $$$${J}''\left({a}\right)=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({ax}\right)}{{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({t}\right)}{\:{t}^{\mathrm{3}} }{dt}=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}.\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{3}\right){sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)} \\ $$$$=−\frac{\mathrm{3}{a}^{\mathrm{2}} \pi}{\mathrm{16}}\:\Rightarrow{J}\left({a}\right)=−\frac{{a}^{\mathrm{4}} \pi}{\mathrm{64}}+{Ka}+{C}\:\: \\ $$$${K}=\mathrm{0}\:\:\:\:\:{C}=\mathrm{0} \\ $$$${G}\left(\mathrm{3}\right)=\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{4}}.\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{5}\right){sin}\left(\frac{\mathrm{5}\pi}{\mathrm{2}}\right)}=\frac{\mathrm{81}\pi}{\mathrm{192}}=\frac{\mathrm{27}\pi}{\mathrm{64}} \\ $$$${J}''\left(\mathrm{1}\right)+{G}\left(\mathrm{3}\right)=\frac{−\pi}{\mathrm{64}}+\frac{\mathrm{27}\pi}{\mathrm{64}}=\frac{\mathrm{13}\pi}{\mathrm{32}} \\ $$
Commented by qaz last updated on 19/Apr/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 20/Apr/21
what mean J(a) sir...
$$\mathrm{what}\:\mathrm{mean}\:\mathrm{J}\left(\mathrm{a}\right)\:\mathrm{sir}… \\ $$

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