Question Number 193278 by Nimnim111118 last updated on 09/Jun/23
$${Please}\:{Help}…!! \\ $$$$\:\:\:\:\underset{\:\:\:\:\mathrm{0}} {\int}^{\:\:\infty} {x}.{e}^{−{x}} .{sinx}.{dx}\: \\ $$$$ \\ $$
Answered by qaz last updated on 09/Jun/23
$$\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}} \mathrm{sin}\:{xdx}=−{im}\int_{\mathrm{0}} ^{\infty} {xe}^{−\left(\mathrm{1}+{i}\right){x}} {dx} \\ $$$$=−{im}\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$−−−−− \\ $$$$\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}} \mathrm{sin}\:{xdx}=−{D}\mathscr{L}\left\{{e}^{−{x}} \mathrm{sin}\:{x}\right\}\left({s}=\mathrm{0}\right)=−{D}\frac{\mathrm{1}}{\left(\mathrm{1}+{s}\right)^{\mathrm{2}} +\mathrm{1}}\left({s}=\mathrm{0}\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}+{s}\right)}{\left(\left(\mathrm{1}+{s}\right)^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\left({s}=\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Nimnim111118 last updated on 10/Jun/23
$${Thank}\:{you}\:{sir}. \\ $$$${But},\:{I}\:{dont}\:{clearly}\:{understand}\:{it}. \\ $$