Help with integral (Rodriquez formula)

In summary, Rodriquez' beta function is a function which is differentiated with each integration, and which includes a 1st term which includes the derivative of $ {x^2-1} $.
  • #1
ognik
643
2
This looked straight forward, but I've gone wrong somewhere ...

Using Rodriquez' formula, find: $ \int_{-1}^{1}{x}^{n}P_n(x) \,dx $ Rodriquez: $ P_n(x)=\frac{1}{2n}\frac{1}{n!}\d{^{n}{({x}^{2}-1)^n}}{{x}^{n}} $

So find $ \int_{-1}^{1}{x}^{n} \frac{1}{2n}\frac{1}{n!}\d{^{n}{({x}^{2}-1)^n}}{{x}^{n}} \,dx $. I'll keep $ \frac{1}{2n}\frac{1}{n!} $ aside for now, and let $ (x^2-1) = f $, and not keep writing in the limits ...

Then, by parts, $ \int {x}^{n} \d{^{n}{f}^n}{{x}^{n}} dx = $ $ {x}^{n} \d{^{n-1}{f}^n}{{x}^{n-1}}|^1_{-1} - \int \d{^{n-1}{f}^n}{{x}^{n-1}} n{x}^{n-1} $ - and repeating by parts n times

I would appreciate knowing how to state better the following arguments around simplifying the problem ...

1st term: After doing 3 Successive differentiations of $ \d{^{n-1}{f}^n}{{x}^{n-1}} $ ALL have a $ ({x}^{2}-1) $ term in them $=(x+1)(x-1)$, therefore I claim all terms with $ \d{^{i}{f}^n}{{x}^{i}} $ in [-1,1] vanish.

I also claim that repeated integration by parts will always have a 1st term which includes $ \d{^{i}{f}^n}{{x}^{i}} $, and to which the above claim applies. So the problem is simplified to the sequence of the integral part only.

Inside the repeated integral will always have 2 parts:
The $x^n$ part: $ x^n $ is differentiated with each integration (always making it the 'u' term for parts). Therefore after n integrations it becomes $n!x^{0}=n! $

The $ \d{^{n}{f^n}}{{x}^{n}} $ part: Because this is always 'v' by parts, each successive integral has a reduced derivative, until finally we are left with just $f^n $

So finally I can find $ \frac{1}{2n}\frac{1}{n!} n! \int_{-1}^{1}f^n \,dx = \frac{1}{2n} \frac{{(x^2-1)}^{n}}{n(n+1)}|^1_{-1}$
I can't see what I have done wrong above?
 
Last edited:
Physics news on Phys.org
  • #2
Hi, I would really appreciate some help with this, I have gone through it a few times with same result, so whatever I have wrong must be some theory that I have wrong, would like to identify that for this and the future ...
 
  • #3
Turns out this is something called a beta function, which was (for whatever reason) a section excluded from the course ...
 

FAQ: Help with integral (Rodriquez formula)

What is the Rodriquez formula used for in integrals?

The Rodriquez formula is a mathematical formula used to solve definite integrals of special functions, such as polynomials and trigonometric functions.

How do you use the Rodriquez formula to solve integrals?

To use the Rodriquez formula, you need to first identify the function you want to integrate and then follow the specific steps outlined in the formula. This usually involves substituting variables and simplifying the expression until you reach a final solution.

Can the Rodriquez formula be used for all types of integrals?

No, the Rodriquez formula is specifically designed for solving definite integrals of special functions. It may not work for all types of integrals, such as indefinite integrals or integrals with complex functions.

Are there any limitations to using the Rodriquez formula?

Yes, the Rodriquez formula can only be used for integrals of special functions and may not work for more complex or advanced integrals. It also requires a good understanding of mathematical concepts and techniques to use effectively.

Can you provide an example of using the Rodriquez formula to solve an integral?

Sure, let's say we want to solve the integral ∫x^3 dx using the Rodriquez formula. We would first substitute x with t and get ∫t^3 dt. Then, we can use the formula ∫t^n dt = (t^(n+1))/(n+1) to get the final solution of (t^4)/4 + C, where C is the constant of integration. Finally, we would substitute t back to x to get the final solution of (x^4)/4 + C.

Similar threads

Replies
16
Views
3K
Replies
2
Views
2K
Replies
5
Views
1K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
6
Views
2K
Back
Top