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Question Number 27410 by math1967 last updated on 06/Jan/18

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{sinxdx}{x}$$

Commented by math1967 last updated on 07/Jan/18

$$Thankyousir$$

Commented by abdo imad last updated on 06/Jan/18

$$letintroducethefonctionf\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}{e}^{-tx}dxwitht⩾0$$$$afterverifyingthatfisderivablewehave$$$$f^{{}^{\prime }}\left(t\right)=-{\int }_0^{\mathrm{\infty }}sinx{e}^{-tx}dx=-Im\left({\int }_0^{\mathrm{\infty }}{e}^{(i-t)x}dx\right)$$$$=-{\left[\frac{1}{i-t}{e}^{(i-t)x}\right]}_{x=0}^{x->\propto }=\frac{1}{i-t}=\frac{i+t}{-1-t^2}=-\frac{t}{1+t^2}-\frac{i}{1+t^2}$$$$\Rightarrow f^{{}^{\prime }}\left(t\right)=-\frac{1}{1+t^2}\Rightarrow f\left(t\right)=\lambda -arctan\left(t\right)andduetofcontinue$$$$\mathrm{\exists }M>0//f\left(t\right)⩽M{\int }_0^{\mathrm{\infty }}{e}^{-tx}dx{=}_{t->\propto so}\frac{M}{t}->0$$$$so\lambda =\frac{\pi }{2}andf\left(t\right)=\frac{\pi }{2}-arctan\left(t\right)$$$${\int }_0^{\mathrm{\infty }}\frac{sinx}{x}=f\left(0\right)=\frac{\pi }{2}.$$$$$$

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