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Question Number 59882 by aliesam last updated on 15/May/19

$${\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x(x^2+1)}dx$$

Commented by maxmathsup by imad last updated on 15/May/19

$$youarewelcome.$$

Commented by Mr X pcx last updated on 15/May/19

$$anotherwaywehave$$$$I={\int }_0^{\mathrm{\infty }}sinx(\frac{1}{x}-\frac{x}{x^2+1})dx$$$$={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}dx-{\int }_0^{\mathrm{\infty }}\frac{xsinx}{x^2+1}dx$$$$=\frac{\pi }{2}-{\int }_0^{\mathrm{\infty }}\frac{xsinx}{x^2+1}dxbut$$$${\int }_0^{\mathrm{\infty }}\frac{xsinx}{x^2+1}dx=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{xsinx}{x^2+1}dx$$$$=\frac{1}{2}Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{xe}^{ix}}{x^2+1}\right)let$$$$w\left(z\right)=\frac{z{e}^{iz}}{z^2+1}thepolesofwareiand$$$$-i\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}w\left(z\right)dz=2i\pi Res(w,i)$$$$Res(w,i)=\frac{i{e}^{-1}}{2i}=\frac{e^{-1}}{2}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}w\left(z\right)dz=2i\pi \frac{e^{-1}}{2}=i\pi e^{-1}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{xsinx}{x^2+1}dx=\frac{\pi }{2}{e}^{-1}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{sinx}{x(1+x^2)}dx=\frac{\pi }{2}-\frac{\pi }{2}{e}^{-1}$$$$=\frac{\pi }{2}(1-\frac{1}{e}).$$

Commented by aliesam last updated on 15/May/19

$$thankyousir$$

Commented by aliesam last updated on 15/May/19

$$thankyousirbrilliantsolution$$

Commented by malwaan last updated on 16/May/19

$$WhatsthefunctionRes?$$

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