About the Legendre differential equation Consider the Legendre differential equation $$ (1-x^2) y'' - 2xy' + n(n+1)y = 0 $$ Then its solution is given by $$ y = c_1 P_n (x) + \text{an infinite series} $$ In fact $y = c_1 P_n (x) + c_2 Q_n (x) $ where $P_n$ is Legendre polynomials and $Q_n $ is Legendre function of the second kind. Here I want to prove that 'an infinite series' above can be written by $c_2 Q_n (x)$ for some constant $c_2$.  
 A: The infinite series you are talking about stems from the recurrence relation that can be derived when solving the Legendre differential equation: plugging in
$$y = \sum_{j=0}^{\infty}a_jx^j $$
into
$$[(1-x^2)y']' + n(n+1)y = 0$$
you get
$$\frac{a_{k+2}}{a_k} = \frac{k(k+1) - n(n+1)}{(k+1)(k+2)}$$
Obviously, this leads to separate solutions with all even or all odd powers of $x$. When $n$ is even, then starting with $a_0 \not= 0$, the series terminates and you get (a multiple of) $P_n$. The same holds for odd $n$ starting with $a_1 \not= 0$.
But when n is even and we start with $a_1 \not= 0$, then the series does not terminate. For instance, for $n=0$, we get
$$\frac{a_{k+2}}{a_k} = \frac{k(k+1)}{(k+1)(k+2)} = \frac{k}{k+2}$$ which clearly leads to (taking $a_1=1$)
$$y= \sum_{k=0}^{\infty}\frac{1}{2k+1}x^{2k+1}$$
Now, as
$$ln(1+x) = \sum_{j=0}^{\infty}\frac{(-1)^{j+1}}{j}x^j$$
it is easy to derive
\begin{equation}
\begin{split}
ln(\frac{1+x}{1-x}) &= ln(1+x) - ln(1-x)
\\ &=\sum_{j=0}^{\infty}\frac{(-1)^{j+1}}{j}x^j + \sum_{j=0}^{\infty}\frac{x^j}{j}
\\ &=\sum_{j=0}^{\infty}\frac{(-1)^{j+1} + 1}{j}x^j
\\ &= 2x + 2\frac{x^3}{3} + 2\frac{x^5}{5} + ...
\end{split}
\end{equation}
which is twice what we are looking for. So
$$Q_0(x) = \frac{1}{2} ln(\frac{1+x}{1-x})$$
Similarly,
$$Q_1(x) = \frac{x}{2} ln(\frac{1+x}{1-x}) - 1$$
And the rest of the $Q_n$ can then be determined by using the recurrence relation
$$(n+1)Q_{n+1}(x) = (2n+1)xQ_n(x) - nQ_{n-1}(x)$$
