Integration Questions

Question Number 91771 by  M±th+et+s last updated on 03/May/20

$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{3}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx} \\$$

Commented by abdomathmax last updated on 03/May/20

$${let}\:{I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{3}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx}\:\:{we}\:{hsve}\:{by}\:{parts} \\$$$${I}\:=\left[−\frac{\mathrm{1}}{{x}}{sin}^{\mathrm{3}} {x}\right]_{\mathrm{0}} ^{\infty} \:+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}}\mathrm{3}{sin}^{\mathrm{2}} {x}\:{cosx}\:{dx} \\$$$$=\mathrm{3}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {x}\:{cosx}}{{x}}{dx} \\$$$$=\frac{\mathrm{3}}{\mathrm{2}}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cosx}\:{sin}\left(\mathrm{2}{x}\right)}{{x}}{dx}\:{we}\:{have} \\$$$${cosx}\:{sin}\left(\mathrm{2}{x}\right)={cosx}\:{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{x}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{cos}\left(−{x}+\frac{\pi}{\mathrm{2}}\right)+{cosx}\left(\mathrm{3}{x}−\frac{\pi}{\mathrm{2}}\right)\right\} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sinx}\:+{sin}\left(\mathrm{3}{x}\right)\right\}\:\Rightarrow \\$$$${I}\:\:=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sinx}\:+{sin}\left(\mathrm{3}{x}\right)}{{x}}{dx} \\$$$$=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{{x}}{dx}\:+\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{3}{x}\right)}{{x}}{dx}\:{but} \\$$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{{x}}{dx}\:=\frac{\pi}{\mathrm{2}} \\$$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{3}{x}\right)}{{x}}{dx}\:=_{\mathrm{3}{x}={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sint}}{\frac{{t}}{\mathrm{3}}}×\frac{{dt}}{\mathrm{3}} \\$$$$=\int_{\mathrm{0}} ^{\infty} \frac{{sint}}{{t}}{dt}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow{I}\:=\frac{\mathrm{3}}{\mathrm{4}}\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{4}}\frac{\pi}{\mathrm{2}} \\$$$$=\frac{\mathrm{3}\pi}{\mathrm{8}}+\frac{\mathrm{3}\pi}{\mathrm{8}}\:=\mathrm{2}×\frac{\mathrm{3}\pi}{\mathrm{8}}\:=\frac{\mathrm{3}\pi}{\mathrm{4}}\:\Rightarrow{I}\:=\frac{\mathrm{3}\pi}{\mathrm{4}} \\$$