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Question Number 58791 by Tawa1 last updated on 30/Apr/19

$$\mathbf{Show}\mathbf{that}:{\int }_0^{\mathrm{\infty }}\frac{\mathbf{sin}\left(\mathbf{x}\right)}{\mathbf{x}}=\frac{\pi }{2}$$

Answered by tanmay last updated on 30/Apr/19

$$ithassometricktosolve...$$$$F\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-ax}sinx}{x}dx$$$$\frac{dF}{da}={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}\times \frac{\partial }{\partial a}\left(e^{-ax}\right)dx$$$$={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}\times e^{-ax}\times -xdx$$$$={\int }_0^{\mathrm{\infty }}-sinx\times e^{-ax}dx$$$$nowlet$$$$p={\int }_0^{\mathrm{\infty }}{e}^{-ax}\times cosxdx$$$$q={\int }_0^{\mathrm{\infty }}{e}^{-ax}\times sinxdx$$$$p+iq={\int }_0^{\mathrm{\infty }}{e}^{-ax}\times e^{ix}dx$$$$p+iq={\int }_0^{\mathrm{\infty }}{e}^{-x(a-i)}dx$$$$=\mid \frac{e^{-x(a-i)}}{-(a-i)}{\mid }_0^{\mathrm{\infty }}=\frac{-1}{-a+i}=\frac{1}{a-i}=\frac{a+i}{a^2+1}$$$$p=\frac{a}{a^2+1}andiq=i\left(\frac{1}{a^2+1}\right)$$$$q=\frac{1}{a^2+1}$$$${\int }_0^{\mathrm{\infty }}-e^{-ax}\times sinxdx=-q=\frac{-1}{a^2+1}$$$${\int }_0^{\mathrm{\infty }}-e^{-ax}\times sinxdx=\frac{-1}{a^2+1}=\frac{dF\left(a\right)}{da}$$$$\int dF\left(a\right)=(-1)\int \frac{da}{a^2+1}$$$$F\left(a\right)=(-1){tan}^{-1}a+c$$$$$$$$asa\to \mathrm{\infty }F\left(a\right)\to 0$$$$0=(-1){tan}^{-1}\left(\mathrm{\infty }\right)+c$$$$0=-\frac{\pi }{2}+c$$$$c=\frac{\pi }{2}$$$$F\left(a\right)=-{tan}^{-1}\left(a\right)+\frac{\pi }{2}$$$$nowF\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-ax}\times \frac{sinx}{x}dx=-{tan}^{-1}\left(a\right)+\frac{\pi }{2}$$$$nowifyouputa=0$$$$weget$$$${\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx=-{tan}^{-1}\left(0\right)+\frac{\pi }{2}$$$${\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx=\frac{\pi }{2}$$$$$$$$$$$$$$$$$$$$$$

Commented by Tawa1 last updated on 30/Apr/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by tanmay last updated on 30/Apr/19

$$thanyou...blessingshowertoall$$

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