- #1

- 104

- 0

## Homework Statement

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

[tex]

Q_n (z) = \frac{1}{2} P_n (z) \ln \left(\frac{z+1}{z-1} \right) + f_{n-1} (z)

[/tex]

implies

[tex]

Q_n (z) = \frac{1}{2} \int \frac{P_n (t) \, dt}{z - t} \quad (n \text{ integer})

[/tex]

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.]

## Homework Equations

Q_n (z) takes the form Q_n (x) in the region -1 < x < 1; x is real:

[tex]

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) \, .

[/tex]

Here's Cauchy's formula:

[tex]

f(z) = \frac{1}{2 \pi i} \oint \frac{f(t) \, dt}{t - z} \, .

[/tex]

## 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: