estimate for highly oscillatory superexponential integral

Question

 I would like to estimate

$\int_{-\pi}^{\pi} e^{i n y} e^{-b e^{c y^{2}}} dy$

to within a RELATIVE error of better than 1%, if possible. Here, $n$ is an integer and $b$ and $c$ are positive. The imaginary component is zero since $\sin ny$ is odd and the rest of the integrand is even.

 It is important to note that $n$ can be up to several hundred, so

this integral is very highly oscillatory. I have tried (and am still trying) to estimate the superexponential part of the integrand as a piecewise polynomial, but this just gives a giant mess.

 A final note: I am trying to obtain a mathematical expression in

terms of $n$, $b$, and $c$. There are several cases to consider (for example, $n=0$ and $n \neq 0$), but I haven't even been able to get the comparatively simple $n = 0$ case so far. Any help would be appreciated.

Answer 1

Using Maple it is possible to obtain

When $n=0$ we have

$$\sum _{k=0}^{\infty }{\frac {-i \left( -1 \right) ^{k}{b}^{k}\sqrt { \pi }{{\rm erf}\left(i\sqrt {k}\sqrt {c}\pi \right)}}{k!\,\sqrt {k} \sqrt {c}}} $$

Answer 2

Let $$ a_n=\int_{-\pi}^{\pi} e^{i n y}\, e^{-b e^{c y^{2}}}dy=2\int_0^{\pi} \cos(n\,y)\, e^{-b e^{c y^{2}}} dy. $$ Let $f(y)=2\,e^{-b e^{c y^{2}}}$. Integration by parts gives $$\begin{align} a_n&=-\frac{1}{n}\int_0^\pi\sin(n\,y)\,f'(y)\,dy\\ &=\frac{(-1)^n}{n^2}\,f'(\pi)-\frac{1}{n^2}\int_0^\pi\cos(n\,x)\,f''(y)\,dy\\ &=\frac{(-1)^n}{n^2}\,f'(\pi)-\frac{(-1)^n}{n^4}\,f'''(\pi)+\frac{1}{n^4}\int_0^\pi\cos(n\,x)\,f^{(4)}(y)\,dy. \end{align}$$ Then $$ \Bigl|a_n-\frac{(-1)^n}{n^2}\,f'(\pi)\Bigr|\le\frac{C_1}{n^4} $$ for some constant $C_1$ (that can be computed explicitly) independent of $n$. For large $n$ this should give a good approximation. If it is not enough, apply integration by parts again to obtain $$ \Bigl|a_n-\Bigl(\frac{(-1)^n}{n^2}\,f'(\pi)-\frac{(-1)^n}{n^4}\,f'''(\pi)\Bigr)\Bigr|\le\frac{C_2}{n^6} $$ for a constant $C_2$.