Welcome to our community

Be a part of something great, join today!

series defined as an integral?

skatenerd

Active member
Oct 3, 2012
114
I've got this funny looking problem for calculus II due tomorrow that I've been stumped on all week. It comes with three parts, and starts by stating:
Define for any \(r\geq0\) (real):
\(\Gamma(r)=\int_{0}^{\infty}{x^r}{e^{-x}}\,dx\)
a. Show that \(\Gamma(0)=1\)
This one was relatively easy. Just plugged in 0 to the r in the integral and got the answer 1.
b. Show that for any \(r\geq0\):
\(\Gamma(r+1)=(r+1)\Gamma(r)\)
When I tried solving for this all I seemed to be able to figure out was to plug the integral that is equal to \(\Gamma(r)\) into the right side of the equation, but from there I really just have no idea what to do with the \(\Gamma(r+1)\).
and if we can get there...
c. Conclude that for any \(n\in N\) (real):
\(\Gamma(n)=n!\)
I feel like solving b might give insight to this problem, but right now all I can say is that I have no idea how a third variable "n" just came into this problem.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
I've got this funny looking problem for calculus II due tomorrow that I've been stumped on all week. It comes with three parts, and starts by stating:
Define for any \(r\geq0\) (real):
\(\Gamma(r)=\int_{0}^{\infty}{x^r}{e^{-x}}\,dx\)
a. Show that \(\Gamma(0)=1\)
This one was relatively easy. Just plugged in 0 to the r in the integral and got the answer 1.
b. Show that for any \(r\geq0\):
\(\Gamma(r+1)=(r+1)\Gamma(r)\)
When I tried solving for this all I seemed to be able to figure out was to plug the integral that is equal to \(\Gamma(r)\) into the right side of the equation, but from there I really just have no idea what to do with the \(\Gamma(r+1)\).
and if we can get there...
c. Conclude that for any \(n\in N\) (real):
\(\Gamma(n)=n!\)
I feel like solving b might give insight to this problem, but right now all I can say is that I have no idea how a third variable "n" just came into this problem.
Hint for b.: use integration by parts on $\Gamma(r+1)$.

Hint for c.: induction.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
b) $\displaystyle \Gamma(r+1)=\int_0^{\infty}x^{r+1}e^{-x}\,dx$

Using IBP, we may define:

$\displaystyle u=x^{r+1}\,\therefore\,du=(r+1)x^r\,dx$

$\displaystyle dv=e^{-x}\,\therefore\,v=-e^{-x}$

and we have...?

edit: pipped at the post! (Tmi)
 
Last edited:

skatenerd

Active member
Oct 3, 2012
114
Thanks for the responses, still a little confused though.
I tried working out the integration by parts for myself and ended up with this:

\(\int_{0}^{\infty}x^{r+1}e^{-x}=-e^{-x}x^{r+1}+(r+1)\int_{0}^{\infty}x^re^{-x}\)

It seems like I'm on the right track but I guess I am overlooking something.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What you have is actually:

$\displaystyle \int_{0}^{\infty}x^{r+1}e^{-x}=\left[-e^{-x}x^{r+1} \right]_0^{\infty}+(r+1)\int_{0}^{\infty}x^re^{-x}\,dx$
 

skatenerd

Active member
Oct 3, 2012
114
Woops! Guess I've never thought to solve that part out with its bounds without solving the whole integral yet.
Anyways, solving that out, I got zero, and that in turn proves what I needed to prove. Thanks a bunch!
One last question, how do you make those longer integral symbols in Latex? They look a lot nicer than these little ones... \(\int f(x) dx\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I use the tags

Code:
$\displaystyle insert LaTeX code here$
 

skatenerd

Active member
Oct 3, 2012
114
Sorry, I just realized I still have no idea how to start part c. Can I just plug in n into \(\Gamma(r)\) ? I still just can't really see any way of relating that integral \(\Gamma(n)\) to n!.
 

chisigma

Well-known member
Feb 13, 2012
1,704
I've got this funny looking problem for calculus II due tomorrow that I've been stumped on all week. It comes with three parts, and starts by stating:
Define for any \(r\geq0\) (real):
\(\Gamma(r)=\int_{0}^{\infty}{x^r}{e^{-x}}\,dx\)
a. Show that \(\Gamma(0)=1\)
This one was relatively easy. Just plugged in 0 to the r in the integral and got the answer 1.
b. Show that for any \(r\geq0\):
\(\Gamma(r+1)=(r+1)\Gamma(r)\)
When I tried solving for this all I seemed to be able to figure out was to plug the integral that is equal to \(\Gamma(r)\) into the right side of the equation, but from there I really just have no idea what to do with the \(\Gamma(r+1)\).
and if we can get there...
c. Conclude that for any \(n\in N\) (real):
\(\Gamma(n)=n!\)
I feel like solving b might give insight to this problem, but right now all I can say is that I have no idea how a third variable "n" just came into this problem.
In order to avoid misunderstanding the 'Gamma Function' is usually defined as...

$\displaystyle \Gamma (r) = \int_{0}^{\infty} x^{r-1}\ e^{- x}\ dx$ (1)

... and the 'Factorial Function' as...

$\displaystyle r! = \int_{0}^{\infty} x^{r}\ e^{- x}\ dx$ (2)

The properties of course are very similar, but it is important don't have confusion. Your example implies the Factorial Function...


Kind regards


$\chi$ $\sigma$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You have stated that:

$\displaystyle n\in\mathbb{N}$

but have (real) after it. I am assuming we are to let n be a natural number instead.

I would begin the proof by induction by demonstrating the validity of the base case:

$\displaystyle \Gamma(1)=1!$

Using the result from part b) we may state:

$\displaystyle \Gamma(0+1)=(0+1)\Gamma(0)$

Using the result from part a) we now have:

$\displaystyle \Gamma(1)=1=1!$

So, the base case is true. Now, state the induction hypotheses $\displaystyle P_k$:

$\displaystyle \Gamma(k)=k!$

From part b) we know $\displaystyle \Gamma(k)=\frac{\Gamma(k+1)}{k+1}$

Now, substitute to finish the proof by induction.
 

skatenerd

Active member
Oct 3, 2012
114
Thanks for all that you guys. And yes I don't know why I put real I did in fact mean natural. I feel like this teacher assumes we all took a class on proofs, but I haven't learned any of these things yet! Guess I'm gonna have to take that class soon.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Induction is sometimes taught in Precalculus, but I suppose it may be optional and up to the discretion of the instructor.

It is a very useful method, and I recommend if you have spare time to give it a look. (Handshake)