Question Number 196557 by BHOOPENDRA last updated on 27/Aug/23
Answered by qaz last updated on 27/Aug/23
![∫_0 ^t e^(−u) sin udu=−ℑ∫_0 ^t e^(−(1+i)u) du=ℑ(1/(1+i))(e^(−(1+i)t) −1) =−(1/2)e^(−t) (sin t+cos t)+(1/2) I=∫_0 ^∞ e^(−t) (1/t)∙(−(1/2)e^(−t) (sin t+cos t)+(1/2))dt =(1/2)∫_0 ^∞ (1/t)(e^(−t) −e^(−2t) (sin t+cos t))dt =(1/2)∫_0 ^∞ 1∙((1/(1+t))−((1+(2+t))/((2+t)^2 +1)))dt =[(1/4)ln((1+2t+t^2 )/(5+4t+t^2 ))−(1/2)arctan (2+t)]_0 ^∞ =(1/4)ln5−(π/4)+(1/2)arctan 2](Q196575.png)
$$\int_{\mathrm{0}} ^{{t}} {e}^{−{u}} \mathrm{sin}\:{udu}=−\Im\int_{\mathrm{0}} ^{{t}} {e}^{−\left(\mathrm{1}+{i}\right){u}} {du}=\Im\frac{\mathrm{1}}{\mathrm{1}+{i}}\left({e}^{−\left(\mathrm{1}+{i}\right){t}} −\mathrm{1}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{t}} \left(\mathrm{sin}\:{t}+\mathrm{cos}\:{t}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} \frac{\mathrm{1}}{{t}}\centerdot\left(−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{t}} \left(\mathrm{sin}\:{t}+\mathrm{cos}\:{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}}\left({e}^{−{t}} −{e}^{−\mathrm{2}{t}} \left(\mathrm{sin}\:{t}+\mathrm{cos}\:{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \mathrm{1}\centerdot\left(\frac{\mathrm{1}}{\mathrm{1}+{t}}−\frac{\mathrm{1}+\left(\mathrm{2}+{t}\right)}{\left(\mathrm{2}+{t}\right)^{\mathrm{2}} +\mathrm{1}}\right){dt} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{4}}{ln}\frac{\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} }{\mathrm{5}+\mathrm{4}{t}+{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\left(\mathrm{2}+{t}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{5}−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:\mathrm{2} \\ $$