# Thread: [SOLVED] Gamma function simple help needed

1. ## [SOLVED] Gamma function simple help needed

I must show, using integration by parts that$\displaystyle \Gamma(x+1)=x\Gamma (x)$ for all positive $\displaystyle x$. I've tried but I don't get the result. The fact that the integral is dt and that the function depends of x is responsible of my fail.
Can you start me up?

2. Originally Posted by arbolis
I must show, using integration by parts that$\displaystyle \Gamma(x+1)=x\Gamma (x)$ for all positive $\displaystyle x$. I've tried but I don't get the result. The fact that the integral is dt and that the function depends of x is responsible of my fail.
Can you start me up?
Let $\displaystyle dv=e^{-t}$

and $\displaystyle t^x=u$

So $\displaystyle v=-e^{-t}$

and $\displaystyle du=\color{red}xt^{x-1}$$\displaystyle \leftarrow\text{ Should look familiar}$

Now notice that your first term dissapears.

3. Let

and

Now notice that your first term dissapears.
I still don't get it. By the way, do making a substitution would allow me to use integration by parts later? If yes, then can you show me how to get to the integration by parts. Thanks.

4. $\displaystyle {\Gamma(p+1)}=\int_{0}^{\infty}x^{p}e^{-x}dx$

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

Let $\displaystyle x^{p}=u, \;\ e^{-x}dx=dv, \;\ du=px^{p-1}dx, \;\ v=-e^{-x}$

Then we get:

$\displaystyle {\Gamma(p+1)}=\left[-x^{p}e^{-x}\right]_{0}^{\infty}-\int_{0}^{\infty}\left(-e^{-x}\right)px^{p-1}dx$

$\displaystyle =p\underbrace{\int_{0}^{\infty}x^{p-1}e^{-x}dx}_{\text{this is gamma}}=p{\Gamma(p)}$

5. Originally Posted by arbolis
I still don't get it. By the way, do making a substitution would allow me to use integration by parts later? If yes, then can you show me how to get to the integration by parts. Thanks.
$\displaystyle \Gamma(x+1)=\int_0^{\infty}t^{x}e^{-t}dt$

So letting $\displaystyle u=t^x\Rightarrow{du=xt^{x-1}}$

and $\displaystyle dv=e^{-t}\Rightarrow{v=-e^{-t}}$

We have

$\displaystyle \Gamma(x+1)=\bigg[-t^{x}e^{-t}\bigg]\bigg|_{0}^{\infty}+\int_0^{\infty}xt^{x-1}e^{-t}dt$

Now for the first term, substitutin zero gives zero, and seeing that $\displaystyle \lim_{t\to\infty}\frac{-t^x}{e^t}=0$

We have

$\displaystyle \Gamma(x+1)=0+\int_0^{\infty}xt^{x-1}e^{-t}dt=x\int_0^{\infty}t^{x-1}e^{-t}dt=x\Gamma(x)\quad\blacksquare$

I have never done this before, but I assume that is how its done.

6. Oh, thanks to both. Now I got it!! But I don't understand one thing.
With $\displaystyle u=t^x$, du is the derivative of u, right? u equals to $\displaystyle e^{xln(t)}$, so the derivative equals $\displaystyle ln(t)e^{xln(t)}=ln(t)*t^x$. How do you get to $\displaystyle du=xt^{x-1}$?

7. You're integrating with respect to t so treat x as a real number: $\displaystyle \frac{d}{dt} t^{n} = nt^{n-1} \: \: n \in \mathbb{R}$

Also, using your method:
$\displaystyle \frac{d}{dt} t^{x} = \frac{d}{dt} e^{x \ln (t)} = e^{x \ln (t)} \cdot (x \ln (t) )' = e^{x \ln (t)} \cdot \frac{x}{t} = t^{x} \cdot \frac{x}{t} = xt^{x-1}$

-- Totally would've answered this thread but too slow .. Gamma ftw!

8. Hello !

Originally Posted by arbolis
Oh, thanks to both. Now I got it!! But I don't understand one thing.
With $\displaystyle u=t^x$, du is the derivative of u, right? u equals to $\displaystyle e^{xln(t)}$, so the derivative equals $\displaystyle ln(t)e^{xln(t)}=ln(t)*t^x$. How do you get to $\displaystyle du=xt^{x-1}$?
You have to be very careful

The notation that is used is not that right.

You should say that u is a function of t.

So this is $\displaystyle \frac{du}{dt}$, not $\displaystyle \frac{du}{dx}$

I think the better way not to be confused is to take n instead of x. So that we're almost sure not to take n as a variable, since we barely use it as a variable

9. You're integrating with respect to t so treat x as a real number:

Also, using your method:
,
You have to be very careful

The notation that is used is not that right.

You should say that u is a function of t.

So this is , not

I think the better way not to be confused is to take n instead of x. So that we're almost sure not to take n as a variable.
Oops!! I got confused! I thought I had to do it with respect to x... how could I understand then, that if $\displaystyle dv=e^{-t}$, then $\displaystyle v=-e^{-t}$!!!
My exam of Calculus II is coming up next 21st of July, so I have all the time to fix such problems of mine. A big thanks to you.

10. Originally Posted by arbolis
, Oops!! I got confused! I thought I had to do it with respect to x... how could I understand then, that if $\displaystyle dv=e^{-t}$, then $\displaystyle v=-e^{-t}$!!!
My exam of Calculus II is coming up next 21st of July, so I have all the time to fix such problems of mine. A big thanks to you.
Although t is a dummy variable (a place holder) in this context as moo said since x is a fixed value and t is the variable you must differentiate in respect to t