Question Number 6081 by LMTV last updated on 12/Jun/16
$$\int{e}^{−{st}} \frac{{t}^{{n}} }{{n}!}{dt}=? \\ $$
Commented by Yozzii last updated on 12/Jun/16
$${Let}\:{I}\left({n}\right)=\frac{\mathrm{1}}{{n}!}\int{e}^{−{st}} {t}^{{n}} {dt}\:\:\left({n}\geqslant\mathrm{0}\right). \\ $$$${Let}\:{u}={st}\Rightarrow{du}={sdt}\Rightarrow\frac{\mathrm{1}}{{s}}{du}={dt}. \\ $$$$\therefore{I}\left({n}\right)=\frac{\mathrm{1}}{{n}!}\int{e}^{−{u}} \left(\frac{{u}}{{s}}\right)^{{n}} ×\frac{\mathrm{1}}{{s}}{du} \\ $$$${I}\left({n}\right)=\frac{\mathrm{1}}{{n}!\left({s}\right)^{{n}+\mathrm{1}} }\left(\int{e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$${See}\:{Qu}.\:\mathrm{6042}\:{for}\:{the}\:{proof}\:{that} \\ $$$$\int{e}^{−{u}} {u}^{{n}} {du}={D}−{n}!{e}^{−{u}} \left(\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{u}^{{n}−{r}} }{\left({n}−{r}\right)!}\right) \\ $$$${where}\:{D}\:{is}\:{a}\:{term}\:{dependent}\:{on}\:{n} \\ $$$${but}\:{independent}\:{of}\:{u}.\:{In}\:{particular} \\ $$$${D}={n}!{C}\:{where}\:{C}\:{is}\:{a}\:{constant}. \\ $$$$\therefore{I}\left({n}\right)=\frac{\mathrm{1}}{{n}!\left({s}\right)^{{n}+\mathrm{1}} }\left({n}!{C}−{n}!{e}^{−{u}} \underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{u}^{{n}−{r}} }{\left({n}−{r}\right)!}\right) \\ $$$${I}\left({n}\right)=\frac{{C}}{\left({s}\right)^{{n}+\mathrm{1}} }−\frac{{e}^{−{u}} }{\left({s}\right)^{{n}+\mathrm{1}} }\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\frac{{u}^{{n}−{r}} }{\left({n}−{r}\right)!}\right) \\ $$$${I}\left({n}\right)={J}−\frac{{e}^{−{st}} }{{s}^{{n}+\mathrm{1}} }\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left({st}\right)^{{n}−{r}} }{\left({n}−{r}\right)!}\:\:\:\left({s}\neq\mathrm{0}\right) \\ $$$$\frac{\mathrm{1}}{{n}!}\int{e}^{−{st}} {t}^{{n}} {dt}={J}−{e}^{−{st}} \left(\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{s}^{−\mathrm{1}−{r}} {t}^{{n}−{r}} }{\left({n}−{r}\right)!}\right) \\ $$$${where}\:{J}=\frac{{C}}{{s}^{{n}+\mathrm{1}} }\:{is}\:{independent}\:{of}\:{t}. \\ $$$$ \\ $$