1. Jun 20, 2010

yungman

I don't understand why I solve the integration in two different ways and get two different answers!!

To find:

$$\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta$$

1) Solve in $\theta$

$$P_1(cos \theta) = cos \theta \;\Rightarrow \; P_1^1(cos \theta)= -sin \theta$$

$$\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int_0^{\pi}sin^2 \theta d \theta = -\frac{\pi}{2}$$

2) Let $$s=cos \theta \;\Rightarrow \; d\theta = \frac{ds}{-sin \theta}$$

$$\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds$$

$$P_1(s)=s \;\Rightarrow P_1^1(s)=1$$

$$\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds = -s|_1^{-1} = 2$$

You see the two method yield two answers!!! I know it should yield the same answer, the book show how to solve the problems in 2) form. Please tell me what did I do wrong.

thanks

Alan

2. Jun 20, 2010

D H

Staff Emeritus
You have the correct expression for the Legendre polynomial P1(s) but not for the associated Legendre function. The correct expression for the associated Legendre functions is given by

$$P_l^m (x) = (-1)^m(1-x^2)^{m/2}\frac{d^m}{dx^m}P_l(x)$$

3. Jun 20, 2010

yungman

Thanks

It must be too late last night. I somehow stuck with the idea that $P_1^1(x) \hbox { is just derivative of } P_1(x)$! I should have gone to sleep instead of posting this!!!

Thanks

Alan