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Question Number 46502 by Ac Kesharwani last updated on 27/Oct/18

$${\int }_0^{\mathrm{\infty }}\left(\frac{\sin \left(\mathbf{x}\right)}{\mathbf{x}}\right)\mathbf{dx}$$

Commented by math khazana by abdo last updated on 27/Oct/18

$$letf\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}{e}^{-tx}dxwitht⩾0wecanprove$$$$thatfiscontinuederivableon[0,+\mathrm{\infty }[andwehave$$$$f^{{}^{\prime }}\left(t\right)=-{\int }_0^{\mathrm{\infty }}{e}^{-tx}sinxdx=-Im\left({\int }_0^{\mathrm{\infty }}{e}^{-tx+ix}dx\right)but$$$${\int }_0^{\mathrm{\infty }}{e}^{(-t+i)x}dx={\left[\frac{1}{-t+i}{e}^{(-t+i)x}\right]}_{x=0}^{+\mathrm{\infty }}=\frac{-1}{-t+i}$$$$=\frac{1}{t-i}=\frac{t+i}{t^2+1}\Rightarrow f^{{}^{\prime }}\left(t\right)=-\frac{1}{1+t^2}\Rightarrow$$$$f\left(t\right)=-arctan\left(t\right)+\lambda but\mathrm{\exists }m>0/$$$$\mid f\left(t\right)\mid ⩽m{\int }_0^{\mathrm{\infty }}{e}^{-tx}dx=\frac{m}{t}\to 0(t\to +\mathrm{\infty })\Rightarrow \lambda =\frac{\pi }{2}$$$$f\left(t\right)=\frac{\pi }{2}-arctan\left(t\right)\Rightarrow f\left(o\right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx=\frac{\pi }{2}$$$$(arctan0=0)$$

Answered by MJS last updated on 27/Oct/18

$$\int \frac{\sin x}{x}dx=\mathrm{Si}\left(x\right)[\mathrm{integral}-\mathrm{sinus},\mathrm{table}\mathrm{of}$$$$\mathrm{values}\mathrm{should}\mathrm{exist}\mathrm{somewhere}$$$$\mathrm{in}\mathrm{the}\mathrm{www}]$$$$\mathrm{Si}\left(\mathrm{\infty }\right)-\mathrm{Si}\left(0\right)=\frac{\pi }{2}$$

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