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Question Number 6042 by FilupSmith last updated on 10/Jun/16
Can you solve the indefinite integral: ∫e^(−u) u^n du
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{indefinite}\:\mathrm{integral}: \\ $$$$\int{e}^{−{u}} {u}^{{n}} {du} \\ $$
Commented by Yozzii last updated on 11/Jun/16
Define I(n)=∫e^(−u) u^n du (n≥0). ∴ Integrating by parts gives I(n)=−e^(−u) u^n −∫−ne^(−u) u^(n−1) du I(n)=−e^(−u) u^n +n∫e^(−u) u^(n−1) du I(n)=−e^(−u) u^n +nI(n−1) I(n)=−e^(−u) u^n −ne^(−u) u^(n−1) +n(n−1)I(n−2) I(n)=−e^(−u) u^n −ne^(−u) u^(n−1) −n(n−1)e^(−u) u^(n−2) +n(n−1)(n−2)I(n−3). I(n)=−e^(−u) (((n!)/((n−0)!))u^(n−0) +((n!)/((n−1)!))u^(n−1) +((n!)/((n−2)!))u^(n−2) +((n!)/((n−3)!))u^(n−3) +...+((n!)/(2!))u^2 )+((n!)/(1!))I(1) I(1)=∫e^(−u) udu=−e^(−u) u+∫e^(−u) du=−e^(−u) u−e^(−u) +C. I(n)=−e^(−u) (((n!)/((n−0)!))u^(n−0) +((n!)/((n−1)!))u^(n−1) +((n!)/((n−3)!))u^(n−3) +...+((n!)/(2!))u^2 +((n!)/(1!))u^1 +((n!)/(0!))u^0 )+D I(n)=D−e^(−u) n!(Σ_(r=0) ^n (u^(n−r) /((n−r)!)))
$${Define}\:{I}\left({n}\right)=\int{e}^{−{u}} {u}^{{n}} {du}\:\left({n}\geqslant\mathrm{0}\right). \\ $$$$\therefore\:{Integrating}\:{by}\:{parts}\:{gives} \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} −\int−{ne}^{−{u}} {u}^{{n}−\mathrm{1}} {du} \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} +{n}\int{e}^{−{u}} {u}^{{n}−\mathrm{1}} {du} \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} +{nI}\left({n}−\mathrm{1}\right) \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} −{ne}^{−{u}} {u}^{{n}−\mathrm{1}} +{n}\left({n}−\mathrm{1}\right){I}\left({n}−\mathrm{2}\right) \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} −{ne}^{−{u}} {u}^{{n}−\mathrm{1}} −{n}\left({n}−\mathrm{1}\right){e}^{−{u}} {u}^{{n}−\mathrm{2}} +{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right){I}\left({n}−\mathrm{3}\right). \\ $$$${I}\left({n}\right)=−{e}^{−{u}} \left(\frac{{n}!}{\left({n}−\mathrm{0}\right)!}{u}^{{n}−\mathrm{0}} +\frac{{n}!}{\left({n}−\mathrm{1}\right)!}{u}^{{n}−\mathrm{1}} +\frac{{n}!}{\left({n}−\mathrm{2}\right)!}{u}^{{n}−\mathrm{2}} +\frac{{n}!}{\left({n}−\mathrm{3}\right)!}{u}^{{n}−\mathrm{3}} +…+\frac{{n}!}{\mathrm{2}!}{u}^{\mathrm{2}} \right)+\frac{{n}!}{\mathrm{1}!}{I}\left(\mathrm{1}\right) \\ $$$${I}\left(\mathrm{1}\right)=\int{e}^{−{u}} {udu}=−{e}^{−{u}} {u}+\int{e}^{−{u}} {du}=−{e}^{−{u}} {u}−{e}^{−{u}} +{C}. \\ $$$${I}\left({n}\right)=−{e}^{−{u}} \left(\frac{{n}!}{\left({n}−\mathrm{0}\right)!}{u}^{{n}−\mathrm{0}} +\frac{{n}!}{\left({n}−\mathrm{1}\right)!}{u}^{{n}−\mathrm{1}} +\frac{{n}!}{\left({n}−\mathrm{3}\right)!}{u}^{{n}−\mathrm{3}} +…+\frac{{n}!}{\mathrm{2}!}{u}^{\mathrm{2}} +\frac{{n}!}{\mathrm{1}!}{u}^{\mathrm{1}} +\frac{{n}!}{\mathrm{0}!}{u}^{\mathrm{0}} \right)+{D} \\ $$$${I}\left({n}\right)={D}−{e}^{−{u}} {n}!\left(\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{u}^{{n}−{r}} }{\left({n}−{r}\right)!}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by FilupSmith last updated on 11/Jun/16
is a D constant?
$$\mathrm{is}\:\mathrm{a}\:{D}\:\mathrm{constant}? \\ $$
Commented by Yozzii last updated on 11/Jun/16
D=n!C where n=constant and C=constant. You get that after substituting the expression for I(1).
$${D}={n}!{C}\:{where}\:{n}={constant} \\ $$$${and}\:{C}={constant}.\:{You}\:{get}\:{that} \\ $$$${after}\:{substituting}\:{the}\:{expression}\:{for}\:{I}\left(\mathrm{1}\right). \\ $$$$ \\ $$
Commented by FilupSmith last updated on 11/Jun/16
I see, thank you!
$$\mathrm{I}\:\mathrm{see},\:\mathrm{thank}\:\mathrm{you}! \\ $$

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