Question Number 139734 by mnjuly1970 last updated on 30/Apr/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\:{nice}\:…{calculus}….. \\ $$$$\:\:\:\:\:\:\:{calculate}\:::\:\: \\ $$$$\:\:\:\:\:\:\mathscr{F}\::=\:\int_{\mathrm{0}} ^{\:\infty} {te}^{−{t}} {sin}^{\mathrm{3}} \left({t}\right){dt}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………………….. \\ $$
Answered by Dwaipayan Shikari last updated on 30/Apr/21
$$\int_{\mathrm{0}} ^{\infty} {te}^{−{t}} \left(\frac{\mathrm{3}}{\mathrm{4}}{sin}\left({t}\right)−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}{i}}\int_{\mathrm{0}} ^{\infty} {te}^{−{t}+{it}} −{te}^{−{t}\left(\mathrm{1}+{i}\right)} −\frac{\mathrm{1}}{\mathrm{8}{i}}\int_{\mathrm{0}} ^{\infty} {te}^{−{t}+\mathrm{3}{it}} −{te}^{−{t}\left(\mathrm{1}+\mathrm{3}{i}\right)} {dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}{i}}.\left(\frac{\mathrm{1}}{\left(\mathrm{1}−{i}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{8}{i}}\left(\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{3}{i}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{3}{i}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}{i}}\left(\frac{\mathrm{4}{i}}{\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{8}{i}}\left(\frac{\mathrm{12}{i}}{\left(\mathrm{10}\right)^{\mathrm{2}} }\right)=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{3}}{\mathrm{200}}=\frac{\mathrm{72}}{\mathrm{200}}=\frac{\mathrm{9}}{\mathrm{25}} \\ $$
Commented by mnjuly1970 last updated on 30/Apr/21
$${thanks}\:{alot}… \\ $$