1. Jun 5, 2010

### yungman

The book gave the integration of a function with the legendre polynomial formula:

$$\int_{-1}^1 f(x)P_n(x)dx = \frac{1}{2^n n!}\int_{-1}^{1} f^n(x)(x^2-1)^n dx$$

It just said the formula can be obtained by repeat using Rodrigues formula and integral by parts but did not go into detail. I want to work out the steps and I got stuck. This is what I have:

Using Rodrigues:

$$\int_{-1}^{1} f(x)P_n(x)dx = \frac{1}{2^n n!}\int_{-1}^{1} f(x)\frac{d^n}{dx^n}[(x^2-1)^n] dx$$

After the first integral by parts:

$$\frac{1}{2^n n!}\int_{-1}^{1} f(x)\frac{d^n}{dx^n}[(x^2-1)^n] dx = \frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 \;-\; \frac{1}{2^n n!}\int_{-1}^{1} f^{(1)}(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n] dx$$

In order for this to continue to the next integration by parts, the following has to be true:

$$\frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 = 0$$

$$\Rightarrow\; [\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 = 0$$

I don't know whether my assumption is correct. If so, I still don't know how it is equal to zero. can anyone give me some guidance.

Last edited: Jun 5, 2010
2. Jun 5, 2010

### LCKurtz

I haven't checked you work, but assuming it is correct, this step:

$$\Rightarrow\; [\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 = 0$$

can be written as

$$[\frac{d^{n-1}}{dx^{n-1}}[(x+1)^n(x-1)^n]]_{-1}^1 = 0$$

Using the product rule differentiating n-1 times I think you will find at least a factor of (x-1)(x+1) in every term of the expansion. That would give you the answer 0.

3. Jun 6, 2010

### yungman

Thanks for you help.