Question

Improper Integral Question. Please see attached picture.

Answer 1

I really do not understand how to do c).

Answer 2

Give me some time to type this... I will use though \[\Gamma(n)=\int_{0}^{\infty}x^{n-1}e^{-x}dx\] (it is the same thing, but you can think of it as using a substituion \(n:=n+1\).. I hope you can see that).

Answer 3

let
\(u=x^{n-1}, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v=-e^{-x}\\
du=(n-1)x^{n-2}dx, \,\,dv=e^{-x}dx\)

Then the integral comes, by integration by parts:
\[=\left[-e^{-x}(x^{n-1})\right]_0^{\infty}-\int_0^{\infty}-e^{-x}(n-1)x^{n-2}dx\\
=\left[-x^{n-1}e^{-x}\right]_0^{\infty}-(n-1)\int_0^{\infty}-e^{-x}x^{n-2}dx=\\
=\left[-x^{n-1}e^{-x}\right]_0^{\infty}+(n-1)\underbrace{\int_0^{\infty}x^{(n-1)-1}e^{-x}dx}_{=\Gamma(n-1)\text{ by definition of the gamma function}}=\\
=\lim_{t\rightarrow \infty}\left.\frac{-x^{n-1}}{e^x}\right|_0^t+(n-1)\Gamma(n-1)\\
=-\lim_{t\rightarrow \infty} t^{n-1}{e^t}+0+(n-1)\Gamma(n-1)\\=
-\lim_{t\rightarrow \infty}\frac{(n-1)!}{e^t} +(n-1)\Gamma(n-1)\\=
-(n-1)!\underbrace{\lim_{t\rightarrow \infty}\frac{1}{e^t}}_{=0}+(n-1)\Gamma(n-1)\\=
(n-1)\Gamma(n-1)\\QED\]

Answer 4

oops i did a typo ... the 2nd line when I used a limit... the \(e^t\) should remain as a denominator in the fraction

Answer 5

now notice that \(\Gamma(n)=(n-1)\Gamma(n-1)\) is the same as \(\Gamma(n+1)=n\Gamma(n)\)

Answer 6

Wow! excellent.

Answer 7

Now in practise... this means that
\(\Gamma(4)=3\Gamma(3)\)
and \(\Gamma(3)=2\Gamma(2)\)
and \(\Gamma(2)=1\Gamma(1)\), and \(\Gamma(1)=1\)
So, \(\Gamma(4)=3*2*1=6\)

In fact, there is te following relationship that exists:
\(\Gamma(n)=(n-1)!\) when \(n\) is an integer! I hope this makes senses from the example above... \(\Gamma(4)=(4-1)!=3!\)

Answer 8

Are you from marianopolis o.O

Answer 9

You're in CEGEP :O?

Answer 10

I used to be, a while ago hehe. I know people who went to Marianopolis though

Answer 11

Oh ok. I'm sure you have a PhD b/c you have some pretty good skills.

Answer 12

no no lol :P, I am in my last year of studies in my math undergrad

Answer 13

They they teach you the gamma function in cegep o_o?

Answer 14

No, it's a question in a bunch of problems too practice in calculus.

Answer 15

May I ask which math undergrad you are doing?

Answer 16

Hm interesting. I only saw this function when I started my 2nd year in statistics

Answer 17

undergrad in stats

Answer 18

I think it is the hardest problem in the package. I doubt I would have this level of difficulty on exams though!

Answer 19

probably not ! Haha I would be very surprised if they tested that at the cegep level

Answer 20

By the way, thanks for the response. That was pretty detailed.

Answer 21

oh you're welcome :) Hehe I love the gamma function, it becomes very useful in stats :)

Answer 22

Good luck with your studies!!