Legendre functions $Q_n(x)$ of the second kind \begin{equation*} Q_n(x)=P_n(x) \int \frac{1}{(1-x^2)\cdot P_n^2(x)}\, \mathrm{d}x \end{equation*} what to do after this step? how can I complete ? I need to reach this formula \begin{equation*} Q_n(x)=\frac{1}{2} P_n(x)\ln\left( \frac{ 1+x}{1-x}\right) \end{equation*}

  • 2
    $\begingroup$ You really should add some $\LaTeX$, as I can't read the formulas $\endgroup$ – Dominic Michaelis Sep 17 '15 at 17:54
  • $\begingroup$ I can not,I do not have more than 10 reputation $\endgroup$ – Jeje Zzz Sep 17 '15 at 18:26
  • $\begingroup$ oh well I start the edit and you can check if I did it correct, I gonna upvote afterwards so you can change things $\endgroup$ – Dominic Michaelis Sep 17 '15 at 19:07
  • $\begingroup$ Do you really need any reputation to edit your own question? $\endgroup$ – mickep Sep 17 '15 at 19:10
  • $\begingroup$ I tried to edit it, but in everyway i read your question the result is just wrong, $Q_1(x)= \frac{x}{2} \ln\left(\frac{1+x}{1-x}\right) -1 $ and your formula gives something different $\endgroup$ – Dominic Michaelis Sep 17 '15 at 19:31

So you want to prove that

$$Q_n(x)= P_n(x) \int^x \frac{1}{(1-x^2) \, \left[P_n(x)\right]^2} \, dx = \frac{1}{2} P_n(x) \ln\left(\frac{1+x}{1-x}\right)$$.

Let's first compute for the case $n=0$, knowing that $P_0(x) = 1$ we would have

$$Q_0(x) = \int^x \frac{1}{(1-x^2)} \, dx = \frac{1}{2} \int \left[\frac{1}{1+x} + \frac{1}{1-x} \right] = \frac{1}{2} \, \ln \left(\frac{1+x}{1-x} \right)$$

where we decomposed the fraction to compute the integral.

Let us now compute for $n=1$, where $P_1(x) = x$.

$$Q_1(x) = x \int^x \frac{1}{(1-x^2)\,x^2} \, dx = \frac{x}{2} \, \ln \left(\frac{1+x}{1-x} \right) - 1$$

where we decomposed the fraction again to compute for the integral.

For $n=2$ we use the same process again, leading us to the result $$Q_2(x) = \frac{1}{2} P_2(x) \ln \left( \frac{1+x}{1-x} \right) - \frac{3x}{2}$$

So it seems that it only works for $n=0$ and that the true formula is

$$Q_n(x) = \frac{1}{2} P_n(x) \ln \left( \frac{1+x}{1-x} \right) - C_n(x)$$

for some value of $C_n(x)$.

As an alternative you could just use the recurrence formulas for $Q_n(x)$ which are

$$(n+1)Q_{n+1}(x) - (2n+1)x \, Q_n(x) + n \, Q_{n-1}(x) = 0$$ $$(2n+1)Q_n(x) = Q'_{n+1}(x) -Q'_{n-1}(x)$$

To show that $$Q_n(x) = \frac{1}{2} P_n(x) \ln \left(\frac{1+x}{1-x}\right) - \frac{2n-1}{n}P_{n-1}(x) - \frac{2n-5}{3(n-1)} P_{n-3}(x) - \cdots$$

If you want to read more about the subject then I suggest you take a look at some mathematical physics books like Mathematical Methods for Physicists by Arfken (Some of my friends say that they dumbed down the seventh edition, but I still use the seventh anyways) or Mathematical Methods in the Physical Sciences by Boas.


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