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Question Number 53967 by maxmathsup by imad last updated on 27/Jan/19

$$1)calculate{A}_t={\int }_0^{\mathrm{\infty }}{e}^{-xt}sinxdxwithx>0$$$$2)byusingFubunitheoremfindthevalueof{\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx.$$

Commented by maxmathsup by imad last updated on 28/Jan/19

$$1)wehave{A}_t=Im\left({\int }_0^{\mathrm{\infty }}{e}^{-xt}{e}^{ix}dx\right)=Im\left({\int }_0^{\mathrm{\infty }}{e}^{(i-t)x}dx\right)$$$${\int }_0^{\mathrm{\infty }}{e}^{(i-t)x}dx={\left[\frac{1}{i-x}{e}^{(i-t)x}\right]}_{x=0}^{x=+\mathrm{\infty }}=-\frac{1}{i-t}=\frac{1}{t-i}=\frac{t+i}{t^2+1}\Rightarrow$$$$A_t=\frac{1}{t^2+1}$$$$2)wehave{\int }_0^{\mathrm{\infty }}{A}_{t}dt={\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+1}={\left[arctant\right]}_0^{+\mathrm{\infty }}=\frac{\pi }{2}andbyfubini$$$${\int }_0^{\mathrm{\infty }}{A}_{t}dt={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{e}^{-xt}sinxdx\right)dt={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{e}^{-xt}dt\right)sinxdx$$$$={\int }_0^{\mathrm{\infty }}\left({[-\frac{1}{x}{e}^{-xt}]}_{t=0}^{t=+\mathrm{\infty }}\right)sinxdx={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx=\frac{\pi }{2}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19

$$B_t={\int }_0^{\mathrm{\infty }}{e}^{-xt}cosxdx$$$$A_t={\int }_0^{\mathrm{\infty }}{e}^{-xt}sinxdx$$$$B_t+{iA}_t={\int }_0^{\mathrm{\infty }}{e}^{-xt}(cosx+isinx)dx$$$$B_t+{iA}_t={\int }_0^{\mathrm{\infty }}{e}^{-xt}.e^{ix}dx={\int }_0^{\mathrm{\infty }}{e}^{-xt+ix}dx$$$$={\int }_0^{\mathrm{\infty }}{e}^{x(-t+i)}dx=\mid \frac{e^{x(-t+i)}}{-t+i}{\mid }_0^{\mathrm{\infty }}$$$$=\mid \frac{e^{-x(t-i)}}{-t+i}{\mid }_0^{\mathrm{\infty }}=\frac{e^{-\mathrm{\infty }(t-i)}-e^0}{-t+i}=\frac{-1}{-t+i}=\frac{1}{t-i}$$$$=\frac{t+i}{t^2+1}=\frac{t}{t^2+1}+i\times \frac{1}{t^2+1}$$$$so{B}_t=\frac{t}{t^2+1}{A}_t=\frac{1}{t^2+1}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19

$$C_t={\int }_0^{\mathrm{\infty }}{e}^{-xt}\frac{sinx}{x}dx$$$$\frac{{dC}_t}{dt}={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial t}\left(\frac{e^{-xt}sinx}{x}\right)dx$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-xt}\times -x\times sinx}{x}dx$$$$=-{\int }_0^{\mathrm{\infty }}{e}^{-xt}sinxdx=-A_t$$$$\frac{{dC}_t}{dt}=-\frac{1}{t^2+1}$$$$-{dC}_t=\frac{dt}{t^2+1}$$$$-C_t={tan}^{-1}\left(t\right)+k$$$$k=-C_t-{tan}^{-1}\left(t\right)$$$$whent\to \mathrm{\infty }{C}_t\to 0and{tan}^{-1}\left(t\right)\to \frac{\pi }{2}$$$$sok=-\frac{\pi }{2}$$$$-C_t={tan}^{-1}\left(t\right)-\frac{\pi }{2}$$$$wehavetofind{\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx$$$$weknowthat$$$$-{\int }_0^{\mathrm{\infty }}{e}^{-xt}\frac{sinx}{x}dx={tan}^{-1}\left(t\right)-\frac{\pi }{2}$$$$$$$$nowputt=0botbside$$$$-{\int }_0^{\mathrm{\infty }}\frac{sinx}{x}={tan}^{-1}\left(0\right)-\frac{\pi }{2}$$$$so{\int }_0^{\mathrm{\infty }}\frac{sinx}{x}=\frac{\pi }{2}proved$$$$$$

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