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Question Number 97492 by 675480065 last updated on 08/Jun/20

Answered by smridha last updated on 08/Jun/20

let ln(x)=−k so we get ∫_0 ^∞ e^(−2021k) k^(2020) dk =((𝚪(2021))/((2021)^(2021) ))[using Laplace Transform]

$$\boldsymbol{{let}}\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)=−\boldsymbol{{k}}\:\boldsymbol{{so}}\:\boldsymbol{{we}}\:\boldsymbol{{get}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\mathrm{2021}\boldsymbol{{k}}} \boldsymbol{{k}}^{\mathrm{2020}} \boldsymbol{{dk}} \\ $$$$=\frac{\boldsymbol{\Gamma}\left(\mathrm{2021}\right)}{\left(\mathrm{2021}\right)^{\mathrm{2021}} }\left[{using}\:\boldsymbol{{Laplace}}\:\boldsymbol{{Transform}}\right] \\ $$

Answered by mathmax by abdo last updated on 08/Jun/20

I =∫_0 ^1 (xlnx)^(2020) dx we do the changement lnx =−t ⇒ I =−∫_0 ^∞ (e^(−t) )^(2020) (−t)^(2020) (−e^(−t) )dt =∫_0 ^∞ e^(−2021 t) t^(2020) dt =_(2021 t =u) ∫_0 ^∞ e^(−u) ((u/(2021)) )^(2020) (du/(2021)) =(1/((2021)^(2021) )) ∫_0 ^∞ u^(2020) e^(−u) du we have Γ(x) =∫_0 ^∞ t^(x−1) e^(−t) dt ⇒ I =(1/((2021)^(2021) )) Γ(2021) =(((2o20)!)/((2021)^(2021) ))

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{xlnx}\right)^{\mathrm{2020}} \:\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{lnx}\:=−\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}\:=−\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{e}^{−\mathrm{t}} \right)^{\mathrm{2020}} \:\left(−\mathrm{t}\right)^{\mathrm{2020}} \:\left(−\mathrm{e}^{−\mathrm{t}} \right)\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{2021}\:\mathrm{t}} \:\mathrm{t}^{\mathrm{2020}} \:\mathrm{dt}\:\:\:=_{\mathrm{2021}\:\mathrm{t}\:=\mathrm{u}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}} \:\left(\frac{\mathrm{u}}{\mathrm{2021}}\:\right)^{\mathrm{2020}} \:\frac{\mathrm{du}}{\mathrm{2021}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2021}\right)^{\mathrm{2021}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{u}^{\mathrm{2020}} \:\mathrm{e}^{−\mathrm{u}} \:\mathrm{du}\:\:\:\mathrm{we}\:\mathrm{have}\:\Gamma\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{x}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}}{\left(\mathrm{2021}\right)^{\mathrm{2021}} }\:\Gamma\left(\mathrm{2021}\right)\:=\frac{\left(\mathrm{2o20}\right)!}{\left(\mathrm{2021}\right)^{\mathrm{2021}} } \\ $$

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