What is the limit of this function as x approaches 1 and -1?

[Pengwuino]

Homework Statement

Basically, what is : $$\left. {\left[ {\ln (1 - x) + \ln (1 + x))(P_{l - 1} (x) - xP_l (x)} \right]} \right|_{x = - 1}^{x = 1}$$

that is, the limits of that function as x->1 and -1. $$P_l(x)$$ is the legendre polynomial of the first kind of order l.

Homework Equations

$$\[ P_l ( - x) = - 1^l P_l (x) \]$$

The Attempt at a Solution

This came from an attempt at Jackson 3.8, creating an expansion for ln(1/sin) in terms of Legendre polynomials. As far as I can tell, the logarithms need to go away before the limit is taken. What I attempted to do was to switch x -> -x in the lower limit which gave me

$$\mathop {\lim }\limits_{x - > 1} (\ln (1 - x) + \ln (1 + x))(P_{l - 1} (x) - xP_l (x)) - \mathop {\lim }\limits_{x - > 1} (\ln (1 - x) + \ln (1 + x))(P_{l - 1} ( - x) + xP_l ( - x))$$

Now, the trick seems to be to be able to turn the right side expression in the 2nd limit to equal the right side expression in the first limit and the logarithms will go away making everything 0. However, I can't seem to do it. The Legendre polynomials are even/odd, but that doesn't seem to do the trick. The negative infront of the x on the second limit ruins me I think. In the actual limit as x-> 1, they're equal but tend to 0, so I need to figure this out before looking at the limit. I'm stumped as to how to do this, possible illegal mathematical operation? Was the switch from x-> -x valid?

 

[lanedance]

have you tried using the inequalities (from a quick wiki to refresh my memory):
$$\frac{x^2-1}{l}P_l'(x) = x P_l(x) -P_{l-1}(x)$$

and
$$P_l'(1) = \frac{l(l+1)}{2}$$

then you get something like a.ln(a) as a->0, though without working it though fully everything seemed to be tending towards zero, not sure how this fits in with your problem

subtitution seems reasonable to me, essentially writing it as
$$\lim_{a\to 1}(ln (1 - x) + ln(1 + x))(P_{l - 1} (x) - xP_l (x))|_{x = - a}^{x = a}$$