Homework $4$: $8\mathrm{.}8,6\mathrm{.}1,6\mathrm{.}2$

Question $1.$ The Gamma function, denoted $(x),$ appears a lot in numerical analysis and probability. It is defined as the following:

$(x)={\displaystyle \underset{0}{\overset{}{}}}{t}^{x-1}{e}^{-t}dt$

Note that the variable of integration is $t.$ One way to solve this integral is by using integration by parts. Part of that process is shown here:

$u=t^{x-1}$Longrightarrowdu$=(x-1)t^{x-2}dt$

$dv=e^{-t}$dtLongrightarrowv$=-e^{-t}$

So$,$

${\displaystyle \underset{0}{\overset{}{}}}{t}^{x-1}{e}^{-t}dt=-t^{x-1}{e}^{-t}{|}_0^{}-{\displaystyle \underset{0}{\overset{}{}}}(x-1)t^{x-2}(-e^{-t})dt||$

Which simplifies to$,$

${\displaystyle \underset{0}{\overset{}{}}}{t}^{x-1}{e}^{-t}dt=-t^{x-1}{e}^{-t}{|}_0^{}+(x-1){\displaystyle \underset{0}{\overset{}{}}}{t}^{x-2}{e}^{-t}dt||$

Lastly, if $(x)={\displaystyle \underset{0}{\overset{}{}}}{t}^{x-1}{e}^{-t}dt,$ then $(x-1)={\displaystyle \underset{0}{\overset{}{}}}{t}^{x-2}{e}^{-t}dt.$ So$,$ we have,

a$)$ Show that $-t^{x-1}{e}^{-t}{|}_0^{}||=0$ for all positive integers $x.$ Then, rewrite the boxed equation above using this result. You may try plugging in small values for $x$ to get a feel for this problem, but you should answer the problem in general. A lot of people will involve L'Hopital's Rule and will write out the logic in words.

$1$