Question Number 6248 by FilupSmith last updated on 20/Jun/16
$$\int_{−\infty} ^{\:\infty} {e}^{−{u}} {u}^{{n}} {du}\:=\:??? \\ $$
Commented by 123456 last updated on 20/Jun/16
$$\mathrm{i}\:\mathrm{think}\:\mathrm{that}\:\mathrm{it}\:\mathrm{diverges} \\ $$
Commented by FilupSmith last updated on 21/Jun/16
$$\mathrm{How}\:\mathrm{can}\:\mathrm{we}\:\mathrm{check}\:\mathrm{this}? \\ $$
Commented by nburiburu last updated on 24/Jun/16
$${It}\:{can}\:{be}\:{solved}\:{by}\:{parts}: \\ $$$$\int{u}.{dv}={u}.{v}−\int{v}.{du} \\ $$$${with} \\ $$$${u}={x}^{{n}} \:\rightarrow\:{du}={n}.{x}^{{n}−\mathrm{1}} \\ $$$${dv}=\:{e}^{−{x}} {dx}\rightarrow\:{v}=−{e}^{−{x}} \\ $$$${since}\:{u}\:{has}\:{a}\:{degree}\:{of}\:{n}\:{it}\:{needs}\:{to}\:{derivate}\:{n}\:{times}. \\ $$$${So},\:{inductively}: \\ $$$$\int{e}^{−{x}} {x}^{{n}} \:{dx}=\:−{e}^{−{x}} {x}^{{n}} −{ne}^{−{x}} {x}^{{n}−\mathrm{1}} −{n}\left({n}−\mathrm{1}\right){e}^{−{x}} {x}^{{n}−\mathrm{2}} −…−\frac{{n}!}{\left({n}−{i}\right)!}\:{e}^{−{x}} {x}^{{n}−{i}} −…−{n}!\:{e}^{−{x}} \\ $$$$\int{e}^{−{x}} {x}^{{n}} {dx}=−{e}^{−{x}} \:\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{\left({n}−{i}\right)!}\:{x}^{{n}−{i}} \\ $$$${Lastly} \\ $$$$\int_{−\infty} ^{+\infty} {e}^{−{x}} \:{x}^{{n}} {dx}=\left(−{e}^{−{x}} \underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{\left({n}−{i}\right)!}{x}^{{n}−{i}} \right)\mid_{−\infty} ^{+\infty} \:= \\ $$$$=−\left[\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\left(\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{\left({n}−{i}\right)!}\:{x}^{{n}−{i}} \right)/{e}^{{x}} \:−\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{\left({n}−{i}\right)!}\:{x}^{{n}−{i}} \right)/{e}^{{x}} \:\right]= \\ $$$$=\:\mathrm{0}\:+\:\infty\:=\:+\infty \\ $$$${The}\:{first}\:{limit}\:{gives}\:\mathrm{0}\:{because}\:{it}\:{is}\:{a}\:{polinomial}\:{divided}\:{exponential}\:{and}\:{using}\:{L}'{hospital}\:{theorem}\:{can}\:{be}\:{demonstrated}. \\ $$$${The}\:{last}\:{limit}\:{gives}\:\infty/\mathrm{0}\:=+\infty \\ $$$$ \\ $$$$ \\ $$