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Question Number 139734 by mnjuly1970 last updated on 30/Apr/21
                                ..... nice ...calculus.....         calculate ::          F := ∫_0 ^( ∞) te^(−t) sin^3 (t)dt=?                        .......................
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\:{nice}\:…{calculus}….. \\ $$$$\:\:\:\:\:\:\:{calculate}\:::\:\: \\ $$$$\:\:\:\:\:\:\mathscr{F}\::=\:\int_{\mathrm{0}} ^{\:\infty} {te}^{−{t}} {sin}^{\mathrm{3}} \left({t}\right){dt}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………………….. \\ $$
Answered by Dwaipayan Shikari last updated on 30/Apr/21
∫_0 ^∞ te^(−t) ((3/4)sin(t)−(1/4)sin(3t))dt  =(3/(8i))∫_0 ^∞ te^(−t+it) −te^(−t(1+i)) −(1/(8i))∫_0 ^∞ te^(−t+3it) −te^(−t(1+3i)) dt  =(3/(8i)).((1/((1−i)^2 ))−(1/((1+i)^2 )))−(1/(8i))((1/((1−3i)^2 ))−(1/((1+3i)^2 )))  =(3/(8i))(((4i)/((1+1)^2 )))−(1/(8i))(((12i)/((10)^2 )))=(3/8)−(3/(200))=((72)/(200))=(9/(25))
$$\int_{\mathrm{0}} ^{\infty} {te}^{−{t}} \left(\frac{\mathrm{3}}{\mathrm{4}}{sin}\left({t}\right)−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}{i}}\int_{\mathrm{0}} ^{\infty} {te}^{−{t}+{it}} −{te}^{−{t}\left(\mathrm{1}+{i}\right)} −\frac{\mathrm{1}}{\mathrm{8}{i}}\int_{\mathrm{0}} ^{\infty} {te}^{−{t}+\mathrm{3}{it}} −{te}^{−{t}\left(\mathrm{1}+\mathrm{3}{i}\right)} {dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}{i}}.\left(\frac{\mathrm{1}}{\left(\mathrm{1}−{i}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{8}{i}}\left(\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{3}{i}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{3}{i}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}{i}}\left(\frac{\mathrm{4}{i}}{\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{8}{i}}\left(\frac{\mathrm{12}{i}}{\left(\mathrm{10}\right)^{\mathrm{2}} }\right)=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{3}}{\mathrm{200}}=\frac{\mathrm{72}}{\mathrm{200}}=\frac{\mathrm{9}}{\mathrm{25}} \\ $$
Commented by mnjuly1970 last updated on 30/Apr/21
thanks alot...
$${thanks}\:{alot}… \\ $$

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