Question

Integrals, yay!

Answer 1

This one has a few parts :)
Consider the function: \[\Gamma(x)=\int\limits_{0}^{infinity}t^{x-1}e^{-1}dt\] a) Find \[\Gamma(1), \Gamma(2)\] b) Integrate by parts to show that, for positive n, \[\Gamma(n+1)=n \Gamma(n)\] c) If n is a positive integer, find a much simpler expression for \[\Gamma(n)\]

Answer 2

So what is the issue?

Answer 3

Also, use `\infty` for \(\infty\).

Answer 4

oh thanks :D well..for part a) do I integrate and then plug the numbers in?

Answer 5

Woops, small typo in your gamma function,\[\large\rm \Gamma(x)=\int\limits\limits_{0}^{\infty}t^{x-1}e^{-t}dt\]But anyway, you plug in the value first, then integrate.\[\large\rm \Gamma(1)=\int\limits\limits\limits_{0}^{\infty}t^{1-1}e^{-t}dt\]

Answer 6

Would that 1-1 by the t just equal t^0
and did you mean e^{-1} ?

Answer 7

I meant e^{-t}
that was your typo :o
This is the Gamma Function, yes?

Answer 8

oh!! yes!
Sorry, I was reading it all wrong.
I thought the x by the t was a t. lol. Wait, let me restart haha

Answer 9

[eating dinner so a bit distracted. foood :>]

Answer 10

mm c:

Answer 11

ok so then it would be t^0

Answer 12

Yes,
I suppose I should have said:
~Plug in the value
~Simplify*
~Integrate

Answer 13

okk xD em..so integration by parts?

Answer 14

No, not for this first one.

Answer 15

\[\large\rm \Gamma(1)=\int\limits_0^{\infty}t^0 e^{-t}dt=\int\limits_0^{\infty} e^{-t}dt\]

Answer 16

:o
\[\Gamma(1)=e^{\infty}-e^{0}\]

Answer 17

oops those are negative ones

Answer 18

Negative, ya?\[\Gamma(1)=-\left[e^{\infty}-e^{0}\right]\]Btw, don't write that on your paper if you're handing it in :)
Your teacher will not like you using infinity as a number lol

Answer 19

Rewrite it as an improper integral if it's something you have to hand in :P\[\large\rm \Gamma(1)=-\left[\lim_{b\to\infty}e^{-b}-e^0\right]\]

Answer 20

I dunno, maybe your teacher doesn't care :D lol

Answer 21

haha yeah i'll do that :) it's a project type thing so it must be most beautiful!! :D

Answer 22

so e^-inft = 0? and e^0 =1

Answer 23

Yes.

Answer 24

-[-1] = 1

Answer 25

\[\large\rm \Gamma(1)=1\]K cool, neeeeext.

Answer 26

\[\Gamma(2)=\int\limits_{0}^{\infty}t^1e^{-1}dt\]

Answer 27

bleeh ^-t on the e

Answer 28

\[\int\limits_{0}^{\infty}te^{-t}dt\] now integration by parts?

Answer 29

mhm

Answer 30

yay! gimme a moment :D

Answer 31

\[= te^{-t}-e^{-t}\]

Answer 32

evaluated from 0 to infty

Answer 33

u = t
dv = e^-t

Answer 34

\[=- te^{-t}-e^{-t}\]

Answer 35

why is the first t negative?

Answer 36

Because uv is the first term.
dv=e^{-t}
v=-e^{-t}

Answer 37

\[=[(-\infty e^{-\infty}-e^{-\infty})-(-0e^{-0}-e^{-0})]\]

Answer 38

0-0-0-1 ?

Answer 39

0-0-0-(-1)

Answer 40

it = 1 too!

Answer 41

\[\large\rm \Gamma(2)=1\]Cool, next!

Answer 42

woowoo
b)integrate by parts to show that, for positive n, gamma(n+1)=n(gamma)n

Answer 43

Mmm k, sounds simple enough.

Answer 44

\[\large\rm \Gamma(n+1)=\int\limits_0^{\infty}t^{n+1-1}e^{-t}dt\]

Answer 45

\[\int\limits_{0}^{t} t^n e^{-t}dt \]

Answer 46

u=t^n
du = nt ?

Answer 47

or nt^{n-1}

Answer 48

ya

Answer 49

the second one?

Answer 50

yes one of those,
ya let's go with the second lol

Answer 51

\[-t^n e^{-t}-\int\limits_{0}^{\infty}-e^{-t}*nt^{n-1}dt\]

Answer 52

k good.

Answer 53

uh so doing by parts again won't really help much.

Answer 54

Pull the negative and n out front,\[\large\rm -t^n e^{-t}+n\color{orangered}{\int\limits\limits_{0}^{\infty}t^{n-1}e^{-t}dt}\]What can we say about this orange thing?

Answer 55

It's super close to what we began with?

Answer 56

It sure is.
It looks like the Gamma Function, but not G(n+1).

Answer 57

oooo because it's n(gamma)n
it's n+1-1 so just n ?

Answer 58

Mmm good yes.\[\large\rm -t^n e^{-t}+n\color{orangered}{\Gamma(n)}\]Just need to show the other part is 0 at both boundaries,
as you did similarly for G(2).

Answer 59

ah. So I evaluate the left of the + from 0 to infinity?

Answer 60

\[-t^n e^{-t} |^\infty_0+n \Gamma(n)\]

Answer 61

mhm

Answer 62

and once we plug those in it all just =0

Answer 63

Part next! :) last one
c) if n is a positive integer, find a much simpler expression for Gamma(n)

Answer 64

...so do i plug n in for x?

Answer 65

I would solve the recurrence relation.

Answer 66

\[\Gamma(n+1)=n \Gamma(n) \\ \text{ with } \Gamma(1)=1 \ \ \text{ divide both sides of } \Gamma(n+1)=n \Gamma(n) \text{ by } \Gamma(n) \\ \frac{\Gamma(n+1)}{\Gamma(n)}= n\]
Then take the product of both sides by from n=1 to n=k

Answer 67

you will see a lot of cancellation

Answer 68

on the left hand side

Answer 69

what what? @freckles

Answer 70

@freckles sorry, I was having a bit of a mini seizure. I just got accepted to the university I want to go to!! :) but now I am back hehe

Answer 71

Why would we set those equal to each other?

Answer 72

did you already do b?

Answer 73

That is where I got the equation from.

Answer 74

Yes we did b

Answer 75

\[\Pi_{n=1}^{k} \frac{\Gamma(n+1)}{\Gamma(n)}=\Pi_{n=1}^{k} n\]

Answer 76

I took the product of both sides

Answer 77

from n=1 to n=k

Answer 78

try simplifying both sides

Answer 79

I'm not sure what "taking the product" means? o.0

Answer 80

you can expand the products on both sides to see what is really happening

Answer 81

have you guys ever did sigma/sum thingys?

Answer 82

instead of adding you multiply in Pi/product thingys

Answer 83

Pi not being the number pi

Answer 84

Pi meaning product here

Answer 85

the E looking things? xD yeah

Answer 86

example right hand side is
\[\prod_{n=1}^{k}n= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdots \cdot (k-1) \cdot k\]

Answer 87

which you can write as k!

Answer 88

try expanding the left hand side a bit

Answer 89

and you should see some really cool stuff happen

Answer 90

wait wait..trying to understand this heh

Answer 91

evaluate the thing we are taking the product of from n=1 to k=n
and just multiply those results

Answer 92

just like you did with sums except we added

Answer 93

\[\Pi = \sum_{}^{}\] type thing?

Answer 94

no products aren't the same as sums
in products we multiply
in sums we add
\[\sum_{i=1}^{3}(3i+5)= (3 \cdot 1+5)+(3 \cdot 2+5)+(3 \cdot 3+5) \\ \prod_{i=1}^{3}(3i+5)=(3 \cdot 1 +5)(3 \cdot 2+5)(3 \cdot 3+5)\]

Answer 95

oh I see o.o

Answer 96

I could have this without introducing that notation
but chances are if you stick in math you will see that notation again

Answer 97

\[\frac{\Gamma(n+1)}{\Gamma(n)}=n\]
So this is what we are trying to solve....
You already proved this equation in b
So this means the following are true:
\[\frac{\Gamma(2)}{\Gamma(1)}=1 \\ \frac{\Gamma(3)}{\Gamma(2)}=2 \\ \frac{\Gamma(4)}{\Gamma(3)}=3 \\ \frac{\Gamma(5)}{\Gamma(4)}=4 \\ \cdots \\ \frac{\Gamma(k)}{\Gamma(k-1)} =k-1 \\ \frac{\Gamma(k+1)}{\Gamma(k)}=k\]

Answer 98

this implies...
\[\frac{\Gamma(2)}{\Gamma(1)} \cdot \frac{\Gamma(3)}{\Gamma(2)} \cdot \frac{\Gamma(4)}{\Gamma(3)} \cdot \frac{\Gamma(5)}{\Gamma(4)} \cdots \frac{\Gamma(k)}{\Gamma(k-1)} \cdot \frac{\Gamma(k+1)}{\Gamma(k)}=1 \cdot 2 \cdot 3 \cdot 4 \cdots (k-1) \cdot k\]