We wish to evaluate:
$$\int_0^{2\pi} P_l^m(\cos\theta)P_{l-1}^m(\cos\theta)\,\text{d}\theta $$
We use the transformation:
$$x = \cos\theta\quad \frac{dx}{d\theta} = -\sin\theta \quad d\theta = \frac{-1}{\sqrt{1-x^2}}dx$$ to obtain$$\int_1^1 \frac{-P_l^m(x)P_{l-1}^m(x)}{\sqrt{1-x^2}}dx = 0$$
Since you're integrating from 1 to 1.
Weird? Yes I thought so too. Seeing as the integral comes from spherical harmoics, I assume you are messing up your limits of integration (should be from $0$ to $\pi$, since that is how the zenith angle is normally defined).
Then, I think you should be working with:
$$\int_0^{\pi} P_l^m(\cos\theta)P_{l-1}^m(\cos\theta)\,\text{d}\theta $$
which after the transformation becomes:
$$I = \int_{-1}^1 \frac{P_l^m(x)P_{l-1}^m(x)}{\sqrt{1-x^2}}dx$$
The Legendre Polynomials satisfy the relationship:
$$P_l^m(-x) = (-1)^{l+m}P_l^m(x)$$
This means that if $l+m$ is even (odd), the Legendre Polynomial is also even (odd). Here we have the two numbers:
$l+m$ and $l+m-1$ One of these two has to be even and the other odd. So one of the polynomials is even and the other is odd. Since $\frac{1}{\sqrt{1-x^2}}$ is even, the integrand as a whole is odd. Thus the integral is also zero.
$$I = 0$$
Reference for parity relationship.
This is an application of the famous saddle-point approximation.
Write
$$ P_n(x)=\int e^{nf(\theta)}d\theta$$
with $f(\theta)=\log(x+\cos(\theta)\sqrt{x^2-1})$.
Saddle-point gives
$$P_n(x)\approx\sqrt{\frac{2\pi}{nf''(\theta_0)}}e^{nf(\theta_0)},$$ where $\theta_0$ is the solution to $f'(\theta_0)=0$.
In your case $\theta_0=0$ and this formula gives what you are looking for.
Best Answer
There must be something wrong. Or maybe I’m wrong.
Using substitution $t=\tan(\varphi/2)$ gives that $$\int_0^\pi \frac{d\,\varphi}{a+b\cos\varphi}=\frac{\pi}{\sqrt{a^2-b^2}}.$$ So in (b) we can get that $P_1(x)=1$ but in (a) we have obviously $P_1(x)=x$.