# series defined as an integral?

#### skatenerd

##### Active member
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
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

Staff member
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!

Last edited:

#### skatenerd

##### Active member
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

Staff member
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
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

Staff member
I use the tags

Code:
$\displaystyle insert LaTeX code here$

#### skatenerd

##### Active member
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
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

Staff member
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
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.