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I am interested in determining the behavior of the following double integral $$ I_N = \int\limits_{0}^{1} \int\limits_{0}^{2 \pi} \Big[ \big( (2x-1)(1+\cos t) -i \sin t \big) x (1-x) \Big]^{N} \, dt \, dx $$ as $N \to \infty$ (one can also write the part depending on $t$ in terms of $e^{\pm it}$ if it helps).

From numerics the integral appears to be exponentially small in $N$ (like $\approx e^{-1.6 N}$, probably up to polynomial factors). However, when one writes the integrand as $f(x,t)^N = e^{N g(x,t)}$, the function $g(x,t)$ has both real and imaginary part, so asymptotic methods such as Laplace's method or stationary phase as I know them are not available out of the box. One source that I found which treats the multidimensional case in some generality is Pemantle and Wilson's paper "Asymptotic expansions of oscillatory integrals with complex phase". In any case, the function $g$ does not have any critical points inside the region of interest, so I am at a loss as how to proceed.

When integrated over $t$, the integrand has its peak at $x = 1/2$, but there are multiple smaller peaks which cancel out each other to some extent, overall it looks like the cancellations are somewhat subtle. Any help would be appreciated.

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    $\begingroup$ Note that $(2x-1) \in [-1,1]$ in your domain, $(1+\cos t)\in [0,2]$ and $\sin t \in [-1,1]$. Your integrand is therefore bounded by $$ | \sqrt{5} x(1-x) |^N $$ Now, the function $x(1-x)$ attains its maximum at $x = 1/2$ with the value $1/4$. So you get that the integrand is bounded by $(5/16)^{N/2} \approx \exp(-0.6N)$. You won't get any help from oscillatory decay. $\endgroup$ Commented Apr 16 at 23:22

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The double integral has a closed form of zero for odd indices and for even index

$I_{2n}= (-1)^n \dfrac{2^{2n+1}}{(n-1/6)!} \dfrac{3^{-3n-1}}{(n+1/6)!} \big( (n-1/2)! \big)^2 \, \pi$

With Gamma function expansion, $I_{2n} \sim (-1)^n \dfrac{2\pi}{3n}\exp{(-3n\log{3})} \big(1-\dfrac{5}{18n} + ...\big) $

Proof sketch follows: Write, with binomial theorem,

$I_n=\int_0^1 (x(1-x))^n \sum_{k=0}^n \binom{n}{k}( 2(2x-1))^k(-2i)^{n-k} \int_0^{2\pi}\cos^{n+k}(t/2) \sin^{n-k}(t/2) dt$

By Beta integral in trig form, the $dt$ integral is solvable in closed form:

$I_n=\int_0^1 (x(1-x))^n \sum_{k=0}^n \binom{n}{k}( 2(2x-1))^k(-2i)^{n-k} \frac{1}{2}(1+(-1)^{n-k})\Gamma(\frac{n+k+1}{2})\Gamma(\frac{n-k+1}{2})/n! $

For $dx$ integral, write $(2x-1)^k = (x-(1-x))^k$ and again use binomial theorem and Beta integral to get a closed form. Wonderfully we find that

$\sum_{m=0}^k \binom{k}{m}(-1)^{k-m}\int_0^1 x^{n+m} (1-x)^{n+k-m} dx = 4^{-1-n}(1+(-1)^k)\Gamma(\frac{k+1}{2})n!/\Gamma(\frac{k+3}{2} + n)$

where the closed-form was generated by Mathematica. Note now that we have factors $(1+(-1)^k) (1+(-1)^{n-k}) $ which implies that both $n$ and $k$ must be even. Collect results thus far...

$I_{2n}=4^{-n}\sum_{k=0}^n\binom{2n}{2k}(-1)^{n-k}\frac{\Gamma(k+1/2)}{\Gamma(2n+k+3/2)} \Gamma(n+k+1/2)\Gamma(n-k+1/2). $

Mathematica can solve in closed-form this sum, which appears as the answer. Readers, please feel free to edit for clarity and a prettier format; too much time went into finding the solution, and not enough for presentation.

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  • $\begingroup$ This is amazing - actually I found my integral by considering a certain alternating combinatorial sum with binomial coefficients etc., which, alas, Mathematica could not simplify to an exact answer. That's why I was looking for asymptotics. $\endgroup$ Commented Apr 22 at 9:21
  • $\begingroup$ Glad to help. It's rare that there's a question on Overflow in my wheelhouse, and I haven't looked on this site for a few years. $\endgroup$ Commented Apr 23 at 3:04
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$\newcommand\ep\varepsilon$Write $$I_n=\int_0^1 dx\int_{-\pi}^\pi dt\, f(t,x)^n,$$ where $f(t,x):=((2x-1)(1+\cos t)-i\sin t)x(1-x)$. The maximum value $1/4$ of $|f(t,x)|$ over all $(t,x)\in[-\pi,\pi]\times[0,1]$ is attained only at the points $(\pm\pi/2,1/2)$. So, it is easy to see that $$I_n=(-i/4+o(1))^n=(-1)^m(1/4+o(1))^{2m} \tag{1}\label{1} $$ for large even $n=2m$. Actually, for even $n$ the value of $I_n$ is real, because $f(-t,x)$ is the complex conjugate of $f(t,x)$, and the bijection $(t,x)\mapsto(-t,x)$ of the set $[-\pi,\pi]\times[0,1]$ preserves the Lebesgue measure.

For odd $n$, we have $I_n=0$, because $f(-t,1-x)=-f(t,x)$ for $(t,x)\in[-\pi,\pi]\times[0,1]$, and the bijection $(t,x)\mapsto(-t,1-x)$ of the set $[-\pi,\pi]\times[0,1]$ preserves the Lebesgue measure.

For an illustration of \eqref{1}, here is the graph $\big\{\big(m,((-1)^m I_{2m})^{1/(2m)}\big)\colon m=1,\dots,100\big\}$ (blue), with a dashed horizontal line at level $1/4$:

enter image description here

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  • $\begingroup$ How do you obtain more information about the $o(1)$ term? Unfortunately, for me $(1/4 + o(1))^{2m}$ is not enough, I'd rather have $h(m)(1/4)^{2m}(1 + o(1))$ for some $h$, possibly with some bounds on the $o(1)$ remainder. $\endgroup$ Commented Apr 17 at 9:15
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    $\begingroup$ @MichałKotowski The integrand is approximately gaussian near the two critical points and this can be used to get an accurate value. $\endgroup$ Commented Apr 17 at 14:38
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    $\begingroup$ @MichałKotowski : I underestimated the complexity caused by the imaginary part. Will try to clarify this. $\endgroup$ Commented Apr 17 at 17:46
  • $\begingroup$ This is indeed what confuses me, since the complex phase itself is not stationary, even though the modulus is. $\endgroup$ Commented Apr 19 at 9:28
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    $\begingroup$ As a matter of fact, it seems to me that the contribution of $(x,t) = (1/2, \pm \pi / 2)$ will not be $\sim (1/4)^n$, but smaller (like $\sim (1/4 \cdot e^{-1/4})^n$, if I made the calculations correctly). If one does Gaussian approximation for $f(x,t) = e^{g(x,t)}$ like Brendan McKay suggested, then the point $(1/2, \pm \pi / 2)$ is not critical for $g$ (it is only critical for the modulus), so after change of variables there will be a correction term (which is why I have $e^{-1/4}$). Then it seems it is no longer straightforward to argue that the main contribution comes from this point only. $\endgroup$ Commented Apr 21 at 18:04

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