# Contour Integration with Legendre Functions

1. Feb 24, 2007

### bigplanet401

1. The problem statement, all variables and given/known data
P_n (z) and Q_n (z) are Legendre functions of the first and second kinds, respectively. The function f is a polynomial in z. Show that
$$Q_n (z) = \frac{1}{2} P_n (z) \ln \left(\frac{z+1}{z-1} \right) + f_{n-1} (z)$$
implies
$$Q_n (z) = \frac{1}{2} \int \frac{P_n (t) \, dt}{z - t} \quad (n \text{ integer})$$
by an application of Cauchy's formula. Be sure to specify the contour.

[Hint: Q_n is many-valued. Cut the z-plane between -1 and 1 along the real axis.]

2. Relevant equations
Q_n (z) takes the form Q_n (x) in the region -1 < x < 1; x is real:
$$Q_n (x) = \frac{1}{2} [Q_n (x + i\epsilon) + Q_n (x - i\epsilon)] = \frac{1}{2} P_n (x) \ln \left( \frac{1+x}{1-x} \right) + f_{n -1}(x) \, .$$

Here's Cauchy's formula:
$$f(z) = \frac{1}{2 \pi i} \oint \frac{f(t) \, dt}{t - z} \, .$$
3. The attempt at a solution

I tried choosing a barbell-shaped contour:

O======O

The "O"s surround the points -1 and +1; the "=====" are a small distance away from the branch cut (x+ie for the upper part, and x-ie for the lower part, where e << 1). The integral over f is zero, since f is entire (it's a polynomial). Got lost after that.

Last edited: Feb 24, 2007
2. Feb 27, 2007

### bigplanet401

(Crickets chirping)