Question Number 129972 by mnjuly1970 last updated on 21/Jan/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:\:{calculus}… \\ $$$$\:\:{evaluation}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {t}^{\mathrm{2}} {e}^{−{t}} {ln}\left({t}\right){dt}=?? \\ $$$$\:\:{solution}: \\ $$$$\:\:\:{f}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} {t}^{\mathrm{2}+{s}} {e}^{−{t}} {dt} \\ $$$$\:\:\:\Omega={f}\:'\left(\mathrm{0}\right)=… \\ $$$$\:\:\:{f}\left({s}\right)=\Gamma\left(\mathrm{3}+{s}\right) \\ $$$$\:\:\:\:{f}\:'\left({s}\right)=\Gamma'\left(\mathrm{3}+{s}\right)=\psi\left(\mathrm{3}+{s}\right)\Gamma\left(\mathrm{3}+{s}\right) \\ $$$${f}\:'\left(\mathrm{0}\right)=\psi\left(\mathrm{3}\right)\Gamma\left(\mathrm{3}\right)=\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\:−\gamma\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}−\mathrm{2}\gamma \\ $$$$\:\:\:\therefore\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {t}^{\mathrm{2}} {e}^{−{t}} {ln}\left({t}\right)=\mathrm{3}−\mathrm{2}\gamma\:… \\ $$$$\:\:\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$
Answered by Ar Brandon last updated on 21/Jan/21
![Ω=∫_0 ^∞ t^2 e^(−t) ln(t)dt=∫_0 ^∞ t^2 e^(−t) ∫_0 ^∞ ((e^(−y) −e^(−ty) )/y)dydt =∫_0 ^∞ ∫_0 ^∞ t^2 {((e^(−t−y) −e^(−ty−t) )/y)}dydt =∫_0 ^∞ ∫_0 ^∞ (1/y){t^2 e^(−t) ∙e^(−y) −t^2 e^(−(y+1)t) }dtdy =∫_0 ^∞ (1/y)Γ(3)e^(−y) dy−∫_0 ^∞ ∫_0 ^∞ (1/y)∙((m^2 e^(−m) )/((y+1)^3 ))dmdy =2∫_0 ^∞ (e^(−y) /y)dy−∫_0 ^∞ ((Γ(3))/(y(y+1)^3 ))dy =2{e^(−y) lny+∫e^(−y) lnydy}_0 ^∞ −2∫_0 ^∞ {(1/y)−(1/(y+1))−(1/((y+1)^2 ))−(1/((y+1)^3 ))}dy =−2γ+[2e^(−y) lny−2lny+2ln(y+1)−(2/(y+1))−(1/((y+1)^2 ))]_0 ^∞ =−2γ+3](https://www.tinkutara.com/question/Q129974.png)
$$\Omega=\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\mathrm{t}} \mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}=\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\mathrm{t}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\mathrm{y}} −\mathrm{e}^{−\mathrm{ty}} }{\mathrm{y}}\mathrm{dydt} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\mathrm{2}} \left\{\frac{\mathrm{e}^{−\mathrm{t}−\mathrm{y}} −\mathrm{e}^{−\mathrm{ty}−\mathrm{t}} }{\mathrm{y}}\right\}\mathrm{dydt} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{y}}\left\{\mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\mathrm{t}} \centerdot\mathrm{e}^{−\mathrm{y}} −\mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\left(\mathrm{y}+\mathrm{1}\right)\mathrm{t}} \right\}\mathrm{dtdy} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{y}}\Gamma\left(\mathrm{3}\right)\mathrm{e}^{−\mathrm{y}} \mathrm{dy}−\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{y}}\centerdot\frac{\mathrm{m}^{\mathrm{2}} \mathrm{e}^{−\mathrm{m}} }{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dmdy} \\ $$$$\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\mathrm{y}} }{\mathrm{y}}\mathrm{dy}−\int_{\mathrm{0}} ^{\infty} \frac{\Gamma\left(\mathrm{3}\right)}{\mathrm{y}\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dy} \\ $$$$\:\:\:\:=\mathrm{2}\left\{\mathrm{e}^{−\mathrm{y}} \mathrm{lny}+\int\mathrm{e}^{−\mathrm{y}} \mathrm{lnydy}\right\}_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left\{\frac{\mathrm{1}}{\mathrm{y}}−\frac{\mathrm{1}}{\mathrm{y}+\mathrm{1}}−\frac{\mathrm{1}}{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{3}} }\right\}\mathrm{dy} \\ $$$$\:\:\:\:=−\mathrm{2}\gamma+\left[\mathrm{2e}^{−\mathrm{y}} \mathrm{lny}−\mathrm{2lny}+\mathrm{2ln}\left(\mathrm{y}+\mathrm{1}\right)−\frac{\mathrm{2}}{\mathrm{y}+\mathrm{1}}−\frac{\mathrm{1}}{\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:\:=−\mathrm{2}\gamma+\mathrm{3} \\ $$
Commented by mnjuly1970 last updated on 21/Jan/21
$${nice}\:{very}\:{nice}\: \\ $$$${thank}\:{you}\:{mr}\:{brandon}… \\ $$
Commented by Ar Brandon last updated on 21/Jan/21
You're welcome Sir��