Convergence of an integral with Legendre polynomials Let's consider the following integral
$$
I(\ell) = \int_{-1}^1 dx P_\ell(x) A(x)
$$
where $ P_\ell(x) $ is the $\ell$-th Legendre polynomial and $A(x) = \frac{1}{1-\lambda x}$ with $0\leq \lambda<1$.
I am interested in estimating the large-$\ell$ behavior of $I(\ell)$. For this purpose, I thought of using the asymptotic expression for the Legendre Polynomials
$$
P_{\ell\gg 1}(x) \simeq  \Re\left[\sqrt{\frac{2}{\pi \ell}} \frac{\left(x+\sqrt{x^2-1}\right)^{\ell+1/2}}{(x^2-1)^{1/4}}\right]
$$
to compute
$$
\tilde{I}(\ell) = \sqrt{\frac{2}{\pi \ell}} \lim_{\epsilon\rightarrow 0}\int_{-1+i \epsilon}^{1+i \epsilon} dx \frac{\left(x+\sqrt{x^2-1}\right)^{\ell+1/2}}{(x^2-1)^{1/4}} A(x)\,.
$$
Then, I will have $I(\ell \gg 1) \simeq \Re\left[\tilde{I}(\ell)\right]$
I can solve this integral numerically and see that $\tilde{I}(\ell)$ is convergent for some values of $\lambda$. However, I can't prove analytically that this integral is convergent since the integrand diverges as $(\epsilon)^{-1/4}$ around $x\sim \pm 1$.
How can I show that $\tilde{I}(\ell)$ is convergent and estimate it?
EDIT : Convergence
Actually, I can show the convergence of the integral by performing the following change of variables
$$
x = \cos{\phi}
$$
after which the integral becomes
$$
I(\ell \gg 1) \simeq \sqrt{\frac{2}{\pi \ell}} \int_{0}^{\pi} d\phi \sqrt{\sin{\phi}} \cos\left[ \phi(\ell+1/2)-\pi/4\right] A(\cos{\phi})\,.
$$
that is clearly convergent for $0\leq \lambda < 1$.
I still don't know how to estimate the high-$\ell$ behaviour
 A: We can use the asymptotic of Legendre Polynomials to evaluate the integral, but we can also use  the Rodrigues' formula:
$$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$$
This approach also allows to find next asymptotic terms in case of needs.
Using this formula,
$$I(n,\lambda)=\frac{1}{2^nn!}\int_{-1}^1\frac{1}{1-\lambda x}\frac{d^n}{dx^n}(x^2-1)^ndx$$
Integrating $n$ times by part,
$$=\frac{(-1)^n\lambda^n}{2^n}\int_{-1}^1\frac{(x^2-1)^n}{(1-\lambda x)^{n+1}}dx=\Big(\frac{\lambda}{2}\Big)^n\int_{-1}^1e^{n\ln\frac{1-x^2}{1-\lambda x}}\frac{dx}{1-\lambda x}$$
Now we can apply the Laplace method for $n\gg1$.
Denoting $\displaystyle f(x)=\ln\frac{1-x^2}{1-\lambda x}$, we find:
$$f'(x)=\frac{\lambda x^2-2x+\lambda}{(1-x^2)(1-\lambda x)}=0\,\text{at}\, x=\frac{1}{\lambda}\pm\sqrt{\frac{1}{\lambda^2}-1}$$
As $x\in(-1;1)$ and $\lambda\in[0;1)$, we have only one root:
$$x_0=\frac{1}{\lambda}-\sqrt{\frac{1}{\lambda^2}-1}=\frac{\lambda}{1+\sqrt{1-\lambda^2}}$$
$$f(x_0)=\ln\frac{1-x_0^2}{1-\lambda x_0}$$
We also find $\displaystyle f''(x_0)=\frac{\lambda^2}{(1-\lambda x_0)^2}-2\frac{1+x_0^2}{(1-x_0^2)^2}=-A$. It is not difficult to check that $A>0\, \big(A\in[2;\infty)$ at $\lambda\in[0;1)\big)$, so the Laplace method is applicable in this case.
Putting the decomposition $\displaystyle f(x)=f(x_0)+f''(x_0)\frac{(x-x_0)^2}{2}$ in our integral, we find the main asymptotic term:
$$I(n,\lambda)\sim\Big(\frac{\lambda}{2}\Big)^n\frac{e^{nf(x_0)}}{1-\lambda x_0}\int_{-1}^1e^{-nA\frac{(x-x_0)^2}{2}}dx\sim\Big(\frac{\lambda}{2}\Big)^n\frac{e^{nf(x_0)}}{1-\lambda x_0}\sqrt\frac{2\pi}{An}$$
Putting explicitly $x_0=\frac{\lambda}{1+\sqrt{1-\lambda^2}}$ in the formula,
$$I(n,\lambda)\sim\left(\frac{\lambda}{2}\frac{1-\frac{\lambda^2}{(1+\sqrt{1-\lambda^2})^2}}{1-\frac{\lambda^2}{1+\sqrt{1-\lambda^2}}}\right)^n\frac{\sqrt\frac{2\pi}{n}}{1-\frac{\lambda^2}{1+\sqrt{1-\lambda^2}}}\left(2\,\frac{1+\frac{\lambda^2}{(1+\sqrt{1-\lambda^2})^2}}{\Big(1-\frac{\lambda^2}{(1+\sqrt{1-\lambda^2})^2}\Big)^2}-\frac{\lambda^2}{\big(1-\frac{\lambda^2}{1+\sqrt{1-\lambda^2}}\big)^2}\right)^{-\frac{1}{2}}\tag{1}$$
This answer can be investigated further (for example, for $\lambda\to 0$ or $\lambda\to 1$).
The easiest check is for $\lambda\to 0$.
At $\lambda=0$ the initial integral converges, and we can expect a simple answer.
Indeed, $x^n$ can be expressed via $x^n=\displaystyle \sum_{k=0}^nc_kP_k(x)$.
But we know that due to orthogonality $\displaystyle \int_{-1}^1P_n(x)P_k(x)=\frac{2}{2n+1} \delta_{kn}$. Using the Rodrigues' formula, we find
$$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}x^{2n}+O(x^{n-1})=\frac{(2n)!}{(n!)^2}\frac{x^n}{2^n}+O(x^{n-1})$$
Using the Stirling' asymptotic for $n!$
$$x^n=\frac{\sqrt{\pi n}}{2^n}P_n(x)+c_{n-1}P_{n-1}+...$$
and
$$\int_{-1}^1\frac{P_n(x)}{1-\lambda x}dx=\int_{-1}^1P_n(x)(1+\lambda x+..+\lambda^nx^n+...)dx=\Big(\frac{\lambda}{2}\Big)^n\frac{2\sqrt{\pi n}}{2n+2}+O(\lambda^{n+1})$$
$$I(\lambda, n)=\sqrt\frac{\pi}{ n}\Big(\frac{\lambda}{2}\Big)^n+o\Big(\sqrt\frac{\pi}{ n}\Big(\frac{\lambda}{2}\Big)^n\Big)$$
This result can be easily derived from the formula (1), leading $\lambda\to 0$.
A: An elaboration on Svyatoslav's answer. Svyatoslav shows that
$$
I(n,\lambda ): = \int_{ - 1}^1 {P_n (t)} \frac{{{\rm d}t}}{{1 - \lambda t}} = \left( {\frac{\lambda }{2}} \right)^n \int_{ - 1}^1 {\exp \left( { - n\log \left( {\frac{{1 - \lambda t}}{{1 - t^2 }}} \right)} \right)} \frac{{{\rm d}t}}{{1 - \lambda t}}.
$$
Now it is convenient to take $\lambda=\tanh \mu$ with an appropriate $\mu\geq 0$. Thus,
$$
I(n,\tanh \mu ) = \left( {\frac{{\tanh \mu }}{2}} \right)^n \int_{ - 1}^1 {{\rm e}^{ - np(t)} } q(t){\rm d}t.
$$
with
$$
p(t) = \log \left( {\frac{{1 - t\tanh \mu }}{{1 - t^2 }}} \right),\quad q(t) = \frac{1}{{1 - t\tanh \mu }}.
$$
The sole saddle point $t_0$ of $p(t)$ on the path of integration is at
$$
t_0  = \tanh \left( {\frac{\mu }{2}} \right).
$$
Applying the saddle point method and simplifying gives
\begin{align*}
& I(n,\tanh \mu )  \sim \left( {\tanh \left( {\frac{\mu }{2}} \right)} \right)^n \sqrt {\frac{\pi }{n}} \frac{{\cosh \left( {\frac{\mu }{2}} \right)}}{{\cosh \mu }}\sum\limits_{k = 0}^\infty  {\frac{{a_k }}{{n^k }}}  \\ & =
\left( {\tanh \left( {\frac{\mu }{2}} \right)} \right)^n \sqrt {\frac{\pi }{n}} \frac{{\cosh \left( {\frac{\mu }{2}} \right)}}{{\cosh \mu }}\left( {1 - \frac{{\cosh ^2 \left( {\frac{\mu }{2}} \right) + 1}}{{8n}} + \frac{{36\cosh ^4 \left( {\frac{\mu }{2}} \right) - 12\cosh ^2 \left( {\frac{\mu }{2}} \right) + 1}}{{128n^2 }} +\ldots } \right)
\end{align*}
as $n\to +\infty$ with any fixed $\mu\geq 0$ and
$$
a_k  = \frac{1}{{\left( {4\cosh \left( {\frac{\mu }{2}} \right)} \right)^{2k} k!}}\left[ {\frac{{{\rm d}^{2k} }}{{{\rm d}t^{2k} }}\left( {\frac{{q(t)}}{{\cosh \mu }}\left( {\cosh ^2 \left( {\frac{\mu }{2}} \right)\frac{{(t - t_0 )^2 }}{{p(t) - p(t_0 )}}} \right)^{k + 1/2} } \right)} \right]_{t = t_0 } .
$$
I suspect that the $a_k$s are polynomials in $\cosh ^2 \left( {\frac{\mu }{2}} \right)$ of degree $k$ with rational coefficients. If that is the case, we must enforce $\mu \ll \log n$ (or, equivalently, $1 - \lambda \gg n^{ - 2}$) in order to obtain useful approximations from the above expansion.
