Question

How to evaluate gamma function at n?

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

But how do we find gamma(1) or something (using this identity)?

Answer 1

If you want gamma(1), you replace n with the value of 1. In general, you evaluate the integral (treating n as a constant and t as a variable). The result will be a function of n

Answer 2

When the integral is evaluated, you will have gotten rid of the t's, so gamma(n) will have only n as a variable.

Answer 3

Yes, but how do you evaluate the integral?

Answer 4

Can't help you in that. I don't remember how you can solve it. Sorry

Answer 5

Newton-Leibniz formula to the rescue !

Answer 6

we ca forget the intergral and consider the defination
\[\Gamma (n)=(n-1)!\]
for any n we can evaluate this

Answer 7

ok, but what about for say n = 1.5?

Answer 8

how ever the improper integral also can be used to
i will prove for
\[\Gamma(1/2)\]

Answer 9

\[\Gamma(\frac{ 1 }{ 2 })=\int\limits _0^{\infty}x^{1/2}e^{-x}dx\]
\[\text{ let u=x^2 then } \implies 2 \int\limits_0^{\infty}e^{-u^2}du=2(\frac{ \sqrt{\pi} }{ 2 })=\sqrt{\pi}\]

Answer 10

for fractions it gets weird but using the fact that e^-x^2 is sqrt of pi we can solve alot of fraction problems

Answer 11

why let u = x^2?

Answer 12

sry i meant x=u^2

Answer 13

wich means u=x^1/2

Answer 14

ok. How did you get from the integral to the next step?

Answer 15

Before you answer that, does it involve anything beyond high school maths?

Answer 16

that requires a new idea not offered in high school math ,its a special function critical to probability

Answer 17

If it's not to difficult, would you care to expand?

Answer 18

^too

Answer 19

Oh, this is the error function...right, I've heard a bit about this.

Answer 20

yes

Answer 21

when you substituted x = u^2, dx = 2u du right?
So then where did the u go in the next step?

Answer 22

you are correct
\[2u du=dx\implies \]
\[\int\limits _0^{\infty}e^{-x} \color{green}{x^{\frac{ -1 }{ 2 }}dx}=\int _0^{\infty} e^{-x}x^{-1/2}u2du=\int _0^{\infty} e^{-u^2}u^{-1}u2du\]
another mistake we have \[x^{1-1/2}=x^{-1/2}\]\

Answer 23

i had made a substitution mistake ealier its -1\2 not 1\2

Answer 24

so th u cancel

Answer 25

Yeah...I'm getting the gist of it.
Can we generalise this to any value of n?

Answer 26

i was wondering about that first we need to consider integers (already fine cos we will use factorial!)then we need to look at numers of the form p/q

i guess with 1/2 we where just lucky in normal cases the x remaines as parrt of the intergral ,but we can always generalize for
\[\Gamma(n/2)\]

Answer 27

So you can do it? What if I gave you like n = 1913/17 or something? or what if n = pi?

Answer 28

its particular values can be expressed in closed form for integer and half-integer arguments, but no simple expressions are known for the values at rational points in general

Answer 29

http://en.wikipedia.org/wiki/Gamma_function

Answer 30

So you can't do exact values for anything that isn't an integer or half integer?

Answer 31

exaclty,as you say it ,mostly rational numbers are approaximated

Answer 32

Interesting...
Thanks heaps for all that by the way

Answer 33

yw,are you really in high school

Answer 34

yeah. Year 11

Answer 35

great,good work and keep it up...sure you wont have problems when you meet up with integrals at college???

Answer 36

Haha thanks!
Actually I'm already doing integrals...it's part of the Australian curriculum