I want to show asymptotics of the following integral involving Legendre Polynomial:

For $0<t<\theta<\frac\pi2$, $$\Big|\int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}} \frac{1-P_n(\cos t)}{t}dt \Big| \leq C \theta^{-\frac12} \cdot \log (n\theta),$$ where $C$ is the constant, $P_n(x)$ is a Legendre Polynomial, $n$ is a positive integer.

I am trying to use Bernstein's inequality for trigonometric polynomials:

Since $P_n(1)=1$ and $|P_n(\cos t)|\leq 1$, $$|P_n(\cos t)-1|=|P_n(\cos t)-P_n(\cos 0)|\leq nt||P_n ||_{\infty}\leq nt.$$

Then I can only get the below estimate but I have no further idea $$\Big|\int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}} \frac{1-P_n(\cos t)}{t}dt \Big| \leq \int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}} \frac{|1-P_n(\cos t)|}{t}dt \leq \int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}} \frac{nt}{t}dt=O(n).$$

Any suggestions are welcome! Thank you for your help!


1 Answer 1


First, suppose that $\frac{1}{n} <\theta <\frac{\pi}{2}$. We split the range of integration into two parts: $0<t<\frac{1}{n}$ and $\frac{1}{n}<t<\theta$, respectively. On the first interval, we use the known limiting relation between the Legendre polynomials and the Bessel function of the first kind $J_0$: \begin{align*} &\int_0^{1/n} {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}\frac{{1 - P_n (\cos t)}}{t}\mathrm{d}t} = \int_0^1 {\frac{1}{{\sqrt \theta + \sqrt {\theta - s/n} }}\frac{{1 - P_n \left( {\cos \left( {\frac{s}{n}} \right)} \right)}}{s}\mathrm{d}s} \\ & \sim \int_0^1 {\frac{1}{{\sqrt \theta + \sqrt {\theta - s/n} }}\frac{{1 - J_0 (s)}}{s}\mathrm{d}s} \le K\int_0^1 {\frac{1}{{\sqrt \theta + \sqrt {\theta - s/n} }} s\, \mathrm{d}s} \le \frac{K}{2}\theta ^{ - 1/2} \end{align*} with a suitable positive constant $K$ and large $n$. On the remaining interval, we have \begin{align*} \left| {\int_{1/n}^\theta {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}\frac{{1 - P_n (\cos t)}}{t}\mathrm{d}t} } \right| & \le \theta ^{ - 1/2} \int_{1/n}^\theta {\left| {\frac{{1 - P_n (\cos t)}}{t}} \right|\mathrm{d}t} \\ & \le 2\theta ^{ - 1/2} \int_{1/n}^\theta \frac{{\mathrm{d}t}}{t} = 2\theta ^{ - 1/2} \log (n\theta ) . \end{align*} Thus, there is a constant $C >0$ such that $$ \left| {\int_0^\theta {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}\frac{{1 - P_n (\cos t)}}{t}\mathrm{d}t} } \right| \le C \theta ^{ - 1/2} \max(1,\log (n\theta )). $$ If $0 <\theta <\frac{1}{n}$, then \begin{align*} & \int_0^\theta {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}} \frac{{1 - P_n (\cos t)}}{t}\mathrm{d}t \le \theta ^{ - 1/2} \int_0^\theta {\frac{{1 - P_n (\cos t)}}{t}\mathrm{d}t} \\ & \le \theta ^{ - 1/2} \int_0^{1/n} {\frac{{1 - P_n (\cos t)}}{t}\mathrm{d}t} \sim \theta ^{ - 1/2} \int_0^1 {\frac{{1 - J_0 (s)}}{s}\mathrm{d}s} \\ & \le K\theta ^{ - 1/2} \int_0^1 s\,\mathrm{d}s = \frac{K}{2}\theta ^{ - 1/2} . \end{align*} In summary, there is a constant $C >0$ such that $$ \left| {\int_0^\theta {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}\frac{{1 - P_n (\cos t)}}{t}\mathrm{d}t} } \right| \le C \theta ^{ - 1/2} \max(1,\log (n\theta )). $$

  • $\begingroup$ Great! Gary, thank you for your nice idea of the Bessel function. I was just wondering how you make $\frac{1-J_0(s)}{s}\leq K s$ from the third integral to the fourth integral? My understanding here is using the series definition of $J_0(s)$. Then we can write $\frac{1-J_0(s)}{s}$ as an alternating series. Since for each term, $0<s<1$, this alternating series can be written as $Ks$ by the technique of sum of geometric series. Is my idea correct here? Thanks again! $\endgroup$
    – user431550
    Nov 28, 2020 at 18:36
  • 1
    $\begingroup$ Yes, I use the local series expansion. In fact, because of the alternation, the first omitted term is an upper bound on $0<s<1$, i.e., $\frac{1-J_0(s)}{s} \leq \frac{s}{4}$ for $0<s<1$. I simplified the final answer a bit. $\endgroup$
    – Gary
    Nov 28, 2020 at 18:47
  • $\begingroup$ Great. Thank you for the clarification! $\endgroup$
    – user431550
    Nov 28, 2020 at 18:58

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