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

$${\int }_0^{\mathrm{\infty }}\frac{\sin x-\sin x^2}{x}=\frac{\pi }{4}$$$${\int }_0^{\mathrm{\infty }}\frac{\cos x-\cos x^2}{x}dx=-\frac{\gamma }{2}$$

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

$${\int }_0^{\mathrm{\infty }}\frac{sin\left(x^{\alpha }\right)}{x}dx{x}^{\alpha }=u\Rightarrow \alpha x^{\alpha -1}=\frac{du}{dx}$$$$=\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}\frac{sin\left(u\right)}{u}du=\frac{\pi }{2\alpha }$$$${\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x}-\frac{sin\left(x^2\right)}{x}dx$$$$=\frac{\pi }{2}-\frac{\pi }{2.2}=\frac{\pi }{4}$$

Answered by mnjuly1970 last updated on 07/Apr/21

$$solution:::::$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)-sin\left(x^2\right)}{x}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x}dx-\{{\int }_0^{\mathrm{\infty }}\frac{sin\left(x^2\right)}{x}dx=\mathbf{\Psi }\}$$$$\therefore \mathit{\varphi }=\frac{\pi }{2}-\mathbf{\Psi };where(\frac{\pi }{2}={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x}dx)$$$$\mathbf{\Psi }\stackrel{⟨x^2=t⟩}{=}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t^{\frac{1}{2}}.t^{\frac{1}{2}}}dt=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt=\frac{\pi }{4}$$$$\therefore \mathit{\varphi }=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$

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

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{sinx}-\sin \left({\mathrm{x}}^2\right)}{\mathrm{x}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}-{\int }_0^{\mathrm{\infty }}\frac{\sin \left({\mathrm{x}}^2\right)}{\mathrm{x}}\mathrm{dx}=\frac{\pi }{2}-{\int }_0^{\mathrm{\infty }}\frac{\sin \left({\mathrm{x}}^2\right)}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{but}{\int }_0^{\mathrm{\infty }}\frac{\sin \left({\mathrm{x}}^2\right)}{\mathrm{x}}\mathrm{dx}{=}_{\mathrm{x}=\sqrt{\mathrm{t}}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{\sqrt{\mathrm{t}}}\frac{\mathrm{dt}}{2\sqrt{\mathrm{t}}}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}=\frac{\pi }{4}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{sinx}-\sin \left({\mathrm{x}}^2\right)}{\mathrm{x}}\mathrm{dx}=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$

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

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\cos \left({\mathrm{x}}^{\mathrm{a}}\right)}{\mathrm{x}}\mathrm{dx}\mathrm{changement}{\mathrm{x}}^{\mathrm{a}}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{t}}^{\frac{1}{\mathrm{a}}}$$$$\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{t}\right)}{{\mathrm{t}}^{\frac{1}{\mathrm{a}}}}{\mathrm{t}}^{\mathrm{a}-1}\mathrm{dt}=\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{a}-\frac{1}{\mathrm{a}}-1}\mathrm{cost}\mathrm{dt}$$$$=\mathrm{Re}\left(\frac{1}{\mathrm{a}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{it}}{\mathrm{t}}^{\mathrm{a}-\frac{1}{\mathrm{a}}-1}\mathrm{dt}\right)\mathrm{we}\mathrm{have}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{a}-\frac{1}{\mathrm{a}}-1}{\mathrm{e}}^{-\mathrm{it}}{=}_{\mathrm{it}=\mathrm{z}}{\int }_0^{\mathrm{\infty }}{(-\mathrm{iz})}^{\mathrm{a}-\frac{1}{\mathrm{a}}-1}{\mathrm{e}}^{-\mathrm{z}}\frac{\mathrm{dz}}{\mathrm{i}}$$$$=(-\mathrm{i}){(-\mathrm{i})}^{\mathrm{a}-\frac{1}{\mathrm{a}}-1}{\int }_0^{\mathrm{\infty }}{\mathrm{z}}^{\mathrm{a}-\frac{1}{\mathrm{a}}-1}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}$$$$={\left({\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}\right)}^{\mathrm{a}-\frac{1}{\mathrm{a}}}\mathrm{\Gamma }(\mathrm{a}-\frac{1}{\mathrm{a}})={\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}(\mathrm{a}-\frac{1}{\mathrm{a}})}.\mathrm{\Gamma }(\mathrm{a}-\frac{1}{\mathrm{a}})\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\mathrm{\Gamma }(\mathrm{a}-\frac{1}{\mathrm{a}}).\cos \left(\frac{\pi }{2}(\mathrm{a}-\frac{1}{\mathrm{a}})\right)\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\cos \left({\mathrm{x}}^2\right)}{\mathrm{x}}\mathrm{dx}$$$$=\mathrm{f}\left(2\right)=\mathrm{\Gamma }(2-\frac{1}{2}).\cos \left(\frac{\pi }{2}(2-\frac{1}{2})\right)=\mathrm{\Gamma }\left(\frac{3}{2}\right).\cos (\pi -\frac{\pi }{4})$$$$=-\frac{\sqrt{2}}{2}\mathrm{\Gamma }\left(\frac{3}{2}\right).....\mathrm{be}\mathrm{continued}....$$

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