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Question Number 196557 by BHOOPENDRA last updated on 27/Aug/23

Answered by qaz last updated on 27/Aug/23

$${\int }_0^t{e}^{-u}\sin udu=-\mathrm{\Im }{\int }_0^t{e}^{-(1+i)u}du=\mathrm{\Im }\frac{1}{1+i}(e^{-(1+i)t}-1)$$$$=-\frac{1}{2}{e}^{-t}(\sin t+\cos t)+\frac{1}{2}$$$$I={\int }_0^{\mathrm{\infty }}{e}^{-t}\frac{1}{t}\cdot (-\frac{1}{2}{e}^{-t}(\sin t+\cos t)+\frac{1}{2})dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1}{t}(e^{-t}-e^{-2t}(\sin t+\cos t))dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}1\cdot (\frac{1}{1+t}-\frac{1+(2+t)}{{(2+t)}^2+1})dt$$$$={[\frac{1}{4}ln\frac{1+2t+t^2}{5+4t+t^2}-\frac{1}{2}\arctan (2+t)]}_0^{\mathrm{\infty }}$$$$=\frac{1}{4}ln5-\frac{\pi }{4}+\frac{1}{2}\arctan 2$$

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