Calculate the asymptotics of the integral $\int\limits_{-\infty }^{\infty }\exp \left [ iw\exp \left ( -x^2 \right ) -\frac{x^2}{2}\right ]dx$ at $w \to \infty $
The difficulty arises right at the beginning; this integral is similar to the integral on the real axis of a function that takes complex values. With $w \to \infty$ this integral cannot be calculated directly. There is a general kind of integral that is used for approximate computation
$$J(\lambda) = \int\limits_a^b f(x)e^{i\lambda \phi(x)}dx, \quad \lambda \rightarrow \infty ,$$
where $f(x)$ and $\phi(x)$ are real functions, and $\lambda$ is a large parameter.
In our case, we can represent the exponent in the integral as $e^{iw\phi(x)}$, where $\phi(x) = e^{-x^2}$, and try to calculate approximately
I can't even approximate it, I also tried to find something here (here), but again nothing worked. Could you help with the asymptotics?
Best Answer
$$I(\omega)\overset{t=e^{-x^2}}{=}\int_0^1\frac{e^{i\omega t}}{\sqrt{-t\ln t}}dt$$ Choosing fixed $a\in(0;1)$, splitting the interval of integration and making the change $t\to1-t$ in the second integral $$I(\omega)=\int_0^a+\int_a^1=\int_0^a\frac{e^{i\omega t}}{\sqrt{-t\ln t}}dt+\int_0^{1-a}\frac{e^{i\omega (1-t)}}{\sqrt{-(1-t)\ln(1-t)}}dt=I_1+I_2\tag{1}$$ $$I_1=\frac1{\sqrt \omega}\int_0^{a\omega}\frac{e^{ix}}{\sqrt{x(\ln\omega-\ln x)}}dx\tag{2}$$ Firsty, we note that for any fixed $0<b<a$ and integrating by part, we get the following estimation: $$\frac1{\sqrt \omega}\int_{b\omega}^{a\omega}\frac{e^{ix}}{\sqrt{x(\ln\omega-\ln x)}}dx=\frac1{i\omega}\frac{e^{i\omega t}}{\sqrt{-t\ln t}}\bigg|_b^a-\frac1{i\omega}\int_b^a\left(\frac1{\sqrt{-t\ln t}}\right)'e^{i\omega t}dt$$ $$=O\left(\frac1\omega\right)\tag{3}$$ Next, fixing $c>0$ $$\Big|\frac1{\sqrt \omega}\int_0^{(1/\omega)^c}\frac{e^{ix}}{\sqrt{x(\ln\omega-\ln x)}}dx\Big|<\frac1{\sqrt \omega}\int_0^{(1/\omega)^c}\frac{dx}{\sqrt{-x\ln x}}=O\left(\frac1{\omega^\frac{c+1}2\sqrt{\ln\,\omega}}\right)\tag{4}$$ Finally, choosing $0<c<1$ $$\frac1{\sqrt \omega}\int_{(1/\omega)^c}^{a\omega}\frac{e^{ix}}{\sqrt{x(\ln\omega-\ln x)}}dx=\frac1{\sqrt{\omega\ln\,\omega}}\int_{(1/\omega)^c}^{a\omega}\frac{e^{ix}}{\sqrt x}\left(1+\frac{\ln x}{2\ln\omega}+...\right)dx$$ $$=\frac1{\sqrt{\omega\ln\,\omega}}\int_0^\infty\frac{e^{ix}}{\sqrt x}dx-\frac1{\sqrt{\omega\ln\,\omega}}\int_{a\omega}^\infty\frac{e^{ix}}{\sqrt x}dx-\frac1{\sqrt{\omega\ln\,\omega}}\int_0^{(1/\omega)^c}\frac{e^{ix}}{\sqrt x}dx+...$$ $$=\frac{\sqrt\pi e^\frac{\pi i}4}{\sqrt{\omega\ln\omega}}+O\left(\frac1{\ln\omega\sqrt{\omega\ln\omega}}\right)\tag{5}$$ We see that the main contribution to $I_1$ comes from (5) $$I_1=\frac1{\sqrt \omega}\int_0^{a\omega}\frac{e^{ix}}{\sqrt{x(\ln\omega-\ln x)}}dx=\frac{\sqrt\pi e^\frac{\pi i}4}{\sqrt{\omega\ln\omega}}+O\left(\frac1{\ln\omega\sqrt{\omega\ln\omega}}\right)\tag{6}$$ Evaluating $I_2$ $$I_2=\frac{e^{i\omega}}{\omega}\int_0^{(1-a)\omega}\frac{e^{-ix}}{\sqrt{1-\frac x\omega}\sqrt{-\ln(1-\frac x\omega)}}$$ $$=\frac{e^{i\omega}}{\omega}\int_0^{(1-a)\omega}\frac{e^{ix}}{\sqrt x}\left(1+\frac x{2\omega}-\frac x{4\omega}+...\right)dx$$ $$I_2=\frac{\sqrt\pi e^{-\frac{\pi i}4}e^{i\omega}}{\sqrt{\omega}}+O\left(\frac1\omega\right)\tag{7}$$ Putting (6) and (7) into (1), $$\boxed{\,\,I(\omega)=\frac{\sqrt\pi e^\frac{\pi i}4}{\sqrt{\omega\ln\omega}}+\frac{\sqrt\pi e^{-\frac{\pi i}4}e^{i\omega}}{\sqrt{\omega}}+O\left(\frac1{\ln\omega\sqrt{\omega\ln\omega}}\right)\,\,}$$ Though the first term is smaller than the second one, it seems that it make sense to keep it - depending on further manipulation. For example, integrating with a smooth function with respect to $\omega$, the second term will bring an additional small factor due to the highly oscillating term $e^{i\omega}$.