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Question Number 184264 by Noorzai last updated on 04/Jan/23

Answered by ARUNG_Brandon_MBU last updated on 04/Jan/23

$$I\left(t\right)={\int }_0^{\mathrm{\infty }}{e}^{-tx}\frac{\sin mx}{x}dx\Rightarrow I^{\prime }\left(t\right)=-{\int }_0^{\mathrm{\infty }}{e}^{-tx}\sin mxdx$$$$\Rightarrow I^{\prime }\left(t\right)=-{\left[\frac{e^{-tx}}{t^2+m^2}(-t\sin mx-m\cos mx)\right]}_0^{\mathrm{\infty }}$$$$=-\frac{m}{t^2+m^2}\Rightarrow I\left(t\right)=-\arctan \left(\frac{t}{m}\right)+C$$$$\underset{t\to +\mathrm{\infty }}{\lim }I\left(t\right)=0=-\frac{\pi }{2}+C\Rightarrow C=\frac{\pi }{2}$$$$\Rightarrow I\left(t\right)=\frac{\pi }{2}-\arctan \left(\frac{t}{m}\right)$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}{e}^x\frac{\sin mx}{x}dx=I(-1)=\frac{\pi }{2}+\arctan \left(\frac{1}{m}\right)$$

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