Im looking for a recurrence formula of type:
$$(\mu-\nu) x P\_\mu^\nu(x) + P\_{\mu-1}^\nu(x)=?, \quad \mu,\nu\in \mathbb R$$
where $P\_\mu^\nu(x)$ is the Legendre function of the first kind (solution to the Legendre differential equation which is regular at the origin). 


My goal is to rewrite the sum in one expression, i.e. $(\mu-\nu) x P\_\mu^\nu(x) + P\_{\mu-1}^\nu(x)= C P\_\alpha^\beta(x) $


Any useful reference, I will be very grateful. Thank you in advance



A recurrence formula for the Legendre function $P_\mu^\nu(x)$
Let's take relation 14.10.3 from the [NIST Handbook](https://dlmf.nist.gov/14.10), which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads
$$
(\mu -\nu ) P\_{\mu }^{\nu } (x) - (2\mu -1)x P\_{\mu -1}^{\nu } (x)
+ (\mu+\nu -1) P\_{\mu -2}^{\nu } (x) =0
$$
We can thus isolate the desired l.h.s.,
$$
(\mu -\nu ) xP\_{\mu }^{\nu } (x) + P\_{\mu -1}^{\nu } (x) =
((2\mu -1)x^2 +1) P\_{\mu -1}^{\nu } (x) - (\mu+\nu -1)x P\_{\mu -2}^{\nu } (x)
$$
The r.h.s. can be consolidated into an expression containing a single Legendre function by also invoking relation 14.10.4 from the [NIST Handbook](https://dlmf.nist.gov/14.10), which, after renaming $\mu \leftrightarrow \nu $ and shifting the new $\mu \rightarrow \mu -2$ reads
$$
(1-x^2 ) \frac{d}{dx} P\_{\mu -2}^{\nu } (x) = 
(\nu-\mu +1) P\_{\mu -1}^{\nu } (x) + (\mu -1)x P\_{\mu -2}^{\nu } (x)
$$
Solving for $P\_{\mu -1}^{\nu } (x)$ and inserting above, we end up with
$$
(\mu -\nu ) xP\_{\mu }^{\nu } (x) + P\_{\mu -1}^{\nu } (x) =
\left[ \frac{(2\mu -1)x^2 +1}{\nu -\mu +1} \left( (1-x^2 ) \frac{d}{dx}
- (\mu -1)x \right) -(\mu +\nu -1)x \right] P\_{\mu -2}^{\nu } (x)
$$


