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


2 Answers 2


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 $\lambda\to 0$.

  • $\begingroup$ Hey, Thank you for this detailed answer. I'll go through all the passages right now but before doing that I was cross-checking your final answer. Are you sure about the $2$ in the denominator in $(\lambda/2)^n$ in your very last equation? I am asking because if I instead put a $1.47$ I get this analytic approximation to follow very well my numerics $\endgroup$
    – apt45
    Oct 20, 2022 at 16:06
  • $\begingroup$ Upvote meanwhile :) $\endgroup$
    – apt45
    Oct 20, 2022 at 16:12
  • $\begingroup$ Oh sorry, your last equation is for $\lambda$ small. I see. That's great. Thanks! $\endgroup$
    – apt45
    Oct 20, 2022 at 16:13
  • $\begingroup$ Thank you :) This factor ($\frac{1}{2^n}$) comes from the definition of Legendre polynomials (we define $P_0=1, P_1=x$, etc.). Also, at $\lambda\to 0$ it is checked by the separate evaluation. I cannot exclude the mistake, though - the evaluation is not difficult, but the expressions are rather bulky... $\endgroup$
    – Svyatoslav
    Oct 20, 2022 at 16:17

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.

  • 1
    $\begingroup$ Really, the ability to find right forms (catching the inherent symmetry) allows to elaborate the problem deeper and to get a general answer! Your solution also explicitly shows that the main multiplier ($\tanh^n\frac{\mu}{2}$) is always $<1$. $\endgroup$
    – Svyatoslav
    Oct 21, 2022 at 8:23

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