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Question Number 137837 by Ñï= last updated on 07/Apr/21

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{x}−\mathrm{sin}\:{x}^{\mathrm{2}} }{{x}}=\frac{\pi}{\mathrm{4}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:{x}−\mathrm{cos}\:{x}^{\mathrm{2}} }{{x}}{dx}=−\frac{\gamma}{\mathrm{2}} \\$$

Answered by Dwaipayan Shikari last updated on 07/Apr/21

$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}^{\alpha} \right)}{{x}}{dx}\:\:\:\:\:{x}^{\alpha} ={u}\Rightarrow\alpha{x}^{\alpha−\mathrm{1}} =\frac{{du}}{{dx}} \\$$$$=\frac{\mathrm{1}}{\alpha}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({u}\right)}{{u}}{du}=\frac{\pi}{\mathrm{2}\alpha} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}−\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}}{dx} \\$$$$=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}.\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\$$

Answered by mnjuly1970 last updated on 07/Apr/21

$$\:\:\:{solution}\:::::: \\$$$$\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right)−{sin}\left({x}^{\mathrm{2}} \right)}{{x}}{dx} \\$$$$\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}−\left\{\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}}{dx}=\boldsymbol{\Psi}\right\} \\$$$$\:\:\:\:\:\:\therefore\:\boldsymbol{\phi}=\:\frac{\pi}{\mathrm{2}}−\boldsymbol{\Psi}\:;\:{where}\:\left(\:\frac{\pi}{\mathrm{2}}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\Psi}\overset{\langle{x}^{\mathrm{2}} ={t}\rangle} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right)}{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} .{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right)}{{t}}{dt}=\frac{\pi}{\mathrm{4}} \\$$$$\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}} \\$$

Answered by mathmax by abdo last updated on 08/Apr/21

$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinx}−\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx}=\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx} \\$$$$\mathrm{but}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx}\:=_{\mathrm{x}=\sqrt{\mathrm{t}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sint}}{\:\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{\mathrm{2}\sqrt{\mathrm{t}}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}\:=\frac{\pi}{\mathrm{4}}\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinx}−\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx}\:=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}} \\$$

Answered by mathmax by abdo last updated on 08/Apr/21

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{x}^{\mathrm{a}} \right)}{\mathrm{x}}\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{x}^{\mathrm{a}} \:=\mathrm{t}\:\mathrm{give}\:\mathrm{x}=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{a}}} \\$$$$\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{t}\right)}{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{a}}} }\mathrm{t}^{\mathrm{a}−\mathrm{1}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}−\mathrm{1}} \:\mathrm{cost}\:\mathrm{dt} \\$$$$=\mathrm{Re}\left(\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{it}} \:\mathrm{t}^{\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}−\mathrm{1}} \mathrm{dt}\right)\:\mathrm{we}\:\mathrm{have} \\$$$$\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{t}^{\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}\:}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{it}} \:=_{\mathrm{it}=\mathrm{z}} \:\int_{\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{iz}\right)^{\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{z}} \:\frac{\mathrm{dz}}{\mathrm{i}} \\$$$$=\left(−\mathrm{i}\right)\left(−\mathrm{i}\right)^{\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}\:−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{z}^{\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{z}} \:\mathrm{dz} \\$$$$=\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}} \:\Gamma\left(\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}\right)\:=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}\left(\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}\right)} \:.\Gamma\left(\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}\right)\:\Rightarrow \\$$$$\mathrm{f}\left(\mathrm{a}\right)=\Gamma\left(\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}\right).\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{a}−\frac{\mathrm{1}}{\mathrm{a}}\right)\right)\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mathrm{dx} \\$$$$=\mathrm{f}\left(\mathrm{2}\right)=\Gamma\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\right).\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\:=\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right).\mathrm{cos}\left(\pi−\frac{\pi}{\mathrm{4}}\right) \\$$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:.....\mathrm{be}\:\mathrm{continued}.... \\$$