Legendre Second Kind: $Q_n(x)$ Functions

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SUMMARY

The discussion centers on the Legendre functions of the second kind, specifically the formula for $Q_n(x)$. The initial expression for $Q_n(x)$ is given as $Q_n(x)=P_n(x) \int \frac{1}{(1-x^2)\cdot P_n^2(x)}\, \mathrm{d}x$. The goal is to derive the alternative expression $Q_n(x)=\frac{1}{2} P_n(x)\ln\left( \frac{ 1+x}{1-x}\right)$. The participants emphasize the need to verify the equivalence of the integral forms to confirm the validity of the proposed relationship.

PREREQUISITES
  • Understanding of Legendre polynomials, specifically $P_n(x)$
  • Familiarity with integral calculus, particularly improper integrals
  • Knowledge of logarithmic functions and their properties
  • Basic grasp of mathematical proofs and verification techniques
NEXT STEPS
  • Study the derivation of Legendre functions, focusing on $Q_n(x)$ and $P_n(x)$
  • Explore integral calculus techniques for evaluating improper integrals
  • Research the properties of logarithmic functions in mathematical analysis
  • Examine mathematical proof strategies to verify function equivalences
USEFUL FOR

Mathematicians, physics students, and researchers working with special functions, particularly those focusing on Legendre functions and their applications in mathematical physics.

jije1112
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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*}
 
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If that were true then, setting the two forms for [itex]Q_(x)[/itex] equal, it would have to be true that
[tex]\int \frac{1}{(1- x^2)P^2_n(x)}dx= \frac{1}{2}ln\left(\frac{1+ x}{1- x}\right)[/tex]

Is that true? I suggest you check your formulas.
 

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