# Why is this integral is super-exponentially small?

Consider the integral $$I_n^{(a,b)} = \int_{-1}^1 (1-x)^a\,(1+x)^b\, P_n(x)\, dx,$$ where $P_n(x)$ is the $n$-th Legendre polynomial. Here's a plot of $|I_n^{(50,20)}|$ for $n=0,\dots,70$: (I just chose the $a,b$ arbitrarily, but the same phenomenon holds for pretty much any choice, and for other polynomials in the Jacobi family as well).

As $n$ ranges from $0$ to $a+b$ the norm of this integral seems to decay like $e^{-n^2}$ (or something faster than exponential anyways). I want to prove a bound to this effect.

What I have tried thus far:

• It's possible to write out the value of the integral in closed form, but it comes out as an alternating sum of huge combinatorial terms which end up cancelling to produce a small number. These seem to be notoriously difficult to analyze and extract meaningful bounds.
• Method of stationary phase / integration by parts. Since the orthogonal polynomials oscillate rapidly for large $n$, and $(1-x)^a(1+x)^b$ is slowly varying (in fact it looks like a bell curve), this seems like a natural approach. Using the trig approximation to the Legendre polynomials one can, eventually, bound the integral (assuming $a\le b$ by a sum of terms resembling $$\frac{1}{n^a} \left| \int_{-1}^1 \frac{d^a}{dx^a}\left[(1+\cos x)^a (1-\cos x)^b\right] \sin x\,e^{i n x}\, dx \right|.$$ The problem here is that now the high order derivatives themselves oscillate quite substantially, and though they can be written down explicitly they contain large alternating terms that are hard to bound. In any case, even if this worked, the bound would only be polynomial in $n$.
• Recursion. By manipulating the integrand and using recurrence properties for the Legendre polynomials you can write an equation $I_n^{(a,b)}$ involving (two) other terms in the sequence. Since the boundary terms are relatively easy to compute, you can try to get a bound by applying the recursion. Here again though, the recursions rely on a lot of cancellations not to blow up, so the bound ends up being very bad.

Basically, I've tried everything I can think of and keep getting thwarted by these annoying cancellation problems. I guess one other technique to try would be steepest descent, but I have little experience with choosing the appropriate contour and suspect I might run into the same issue. Any tips from the experts would be most welcome. Thanks!

• Did you know that $P_n(\cos(x))\rightarrow J_0(nx)+O(n^{-1})$ as $n\rightarrow\infty$ Apr 24, 2017 at 23:52
• For such large a,b the first two terms are extraordinarily tiny away from $-1, 1$... but not that small. Rodrigues' formula for $P_n$ seems clearly very small near $1,-1$. Perhaps breaking the integral up, and doing some kind of local approximation near the end points vs center, will do what you want? Apr 24, 2017 at 23:55
• @tired yes. But integrating even simple things against $J_0$ leads nowhere nice, e.g. $\int_0^\pi J_0(n x) x^a\,dx = \frac{\pi^{a+1}}{1+a} \, _1F_2\left(\frac{a}{2}+\frac{1}{2};1,\frac{a}{2}+\frac{3}{2};-\frac{1}{4} n^2 \pi ^2\right)$. If I knew how to read off bounds for the hypergeometric PFQ function I'd be done since the original integral can be expressed as such.
– nth
Apr 25, 2017 at 2:17
• The decay suggests that "morally" this is integral is like Fourier transform of a Gaussian. I think you can extract a proof from this idea since for sufficiently large $a,b,n$, the first part converges to (shifted, rescaled) $e^{-x^2}$ and $P_n(x)$ looks locally like a cosine wave over the not-exponentially-small part of the support of the Gaussian.
– nth
Apr 25, 2017 at 16:42
• ok i think we need $a+b\sim n$ to make this approximation work (establish an asymptotic equivalence) which means that in this sense the decay is indeed linear Apr 25, 2017 at 19:15

This proof will derive an 'envelope' function $E$ based on an asymptotic expansion for $I_n^{(a,b)}.$ The asymptotic expansion is good for $n \lesssim 0.2(a+b).$ It will take a lot more work to get an asymptotic expansion for larger $n/(a+b),$ but numerical work suggests that the one derived is an upper bound.
Use the known expansion for the Legendre polynomials, $$(A)\quad P_n(x)=2^{-n} \sum_{k=0}^n \binom{n}{k}^2 (x-1)^{n-k}(x+1)^k .$$ Insert into the integral and use the beta integral $$\int_{-1}^{1}(1+x)^{\mu-1}(1-x)^{\nu-1}\,dx = 2^{\mu + \nu -1} \frac{\Gamma(\mu)\Gamma(\nu)}{\Gamma(\nu + \mu + 1)}.$$ With simplification we get the identity $$I_n^{(a,b)}=\underbrace{2^{a + b +1}\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+n+2)}}_{G_n(a,b)}\;\underbrace{ \sum_{k=0}^n\binom{n}{k}^2(-1)^{n-k} \big(1+b\big)_k \big(1+a\big)_{n-k}}_{S_n(a,b)}$$ where the Pochhammer symbol $\big(a\big)_k = \Gamma(a+k)/\Gamma(a)$ has been used.
$$S_n(a,b)=n!^2\sum_{k=0}^n (-1)^{n-k} \frac{\big(1+b\big)_k}{k!^2} \frac{\big(1+a\big)_{n-k}}{(n-k)!^2}$$ $$=n!^2\,[x^n]\sum_{k=0}^\infty \frac{(-x)^k}{k!^2}\big(1+a\big)_k \sum_{k=0}^\infty \frac{x^m}{m!^2}\big(1+b\big)_m =$$ $$=n!^2 [x^n] \big( e^{-x} {}_1F_1(-a,1,x) \big) \big( e^{x} {}_1F_1(-b,1,-x) \big) .$$ In the previous string of calculations $[x^n]$ means 'coefficient of' and the statement follows by recognizing the single sum as what comes from a Cauchy product of 2 power series. In the next line the power series are written in terms of a confluent hypergoemetric series, and furthermore Kummer's transform has been used. Now assume $a$ and $b$ are large and use the asymptotic relations $${}_1F_1(-a,1,x) =L_a(x) \sim e^{x/2}J_0(2\sqrt{ax})$$ $${}_1F_1(-b,1,-x) =L_b(-x) \sim e^{-x/2}I_0(2\sqrt{ax})$$ where $J_0$ and $I_0$ are Bessel functions. Thus $$S_n(a,b)\sim [x^n] n!^2\sum_{k=0}^\infty (-1)^k \frac{(a\,x)^k}{k!^2} \sum_{m=0}^\infty \frac{(b\,x)^m}{m!^2}$$ $$=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}^2 b^k a^{n-k} = (-1)^n(b+a)^n P_n(\frac{a-b}{a+b}).$$ where formula (A) has again been used. In what follows let $\cos \theta = (a-b)/(a+b)$ since the ratio has a magnitude $<1.$
The neat thing about this calculation is that the interpretation of 'coefficient of' never had to enter the complex plane as a contour integral around the origin. Note: there are better approximations of the Laguerre functions in which the latter interpretation would be needed, but this simple treatment is sufficient for what I want to show. Use the well-known asymptotic relationship for the Legendre polynomials, $$P_n(\cos \theta) = \sqrt{\frac{2}{\pi n\sin \theta}} \cos((n+\tfrac{1}{2})\theta - \pi/4) + O(n^{-3/2}).$$ The envelope function is derived by setting any oscillatory terms dependent on $n$ equal to 1, i.e., $$E=env(I_n^{(a,b)}) \sim f \frac{(a+b)^n}{\sqrt{n}\, \Gamma(a+b+n+2)} \, , \, f = 2^{a+b+1}\Gamma(a+1)\,\Gamma(b+1) \,\sqrt{\frac{2}{\pi n\sin \theta}}$$ Now if a function $F$ had a merely exponential decay, say, $F=\exp(-c\,n),$ then $d/dn \, \log{F} = -c.$ For the given $E$ it is easy to show that the result of this procedure results in an increasing function of $n$ with a negative coefficient; in other words, superlinear decay. $$\frac{d}{dn} \, \log{E} = \frac{d}{dn} \Big( n \log(a+b) -\tfrac{1}{2} \log n - \log{\Gamma(a+b+n+2)} \Big)$$ Use the asymptotics of the digamma function, $$\psi(x)= \frac{d}{dx} \log{\Gamma(x)} \sim \log{x} - \frac{1}{2x} + ...$$ and some simplification to conclude that $$\frac{d}{dn} \, \log{E} = -\log{(1+\frac{n+2}{a+b})} .$$ This formula is sufficient to prove the faster-than-linear decay hypothesis, at least for small $n/(a+b).$ It tends to underestimate the decay for larger $n.$