Find the leading behaviour

Question

Find the leading behaviour of given integral as $x$ tends to infinity$$ I(x)=\int_0^1e^{ixy^2}\cosh y^2 dy$$ Here, to use the method of stationary phase, $y=0$ is a minimum point of $y^2$ and $y''(0)\neq0$, if we take $f(y)=\cosh y^2$ and $g(y)=t^2$, is it right to write leading behaviour as, $I(x)$~$f(0)e^{ixg(0)+\pi/4}\sqrt{\frac{\pi}{2x}}$?

Answer 1

(I am not sure what $g(0)$ refers to in your proposed solution, so I give here a self contained answer)

As you have pointed out the integral for $x\to\infty$ is dominated by the stationary point at $y=0$. In order to obtain the asymptotic expansion, we have to deform the integration contour to a contour with constant phase 0, that is along a line $y =e^{i\pi/4} t$ with $t\geq 0$.

There is an additional contribution coming from the end point (the integration contour has to come back from $y\to e^{i\pi/4}\infty$ to $y=1$) along a line of constant phase $e^{ix}$. This line is parameterized by $\xi\geq$ with $y= \cosh\xi+i \sinh\xi.$

With this procedure, we obtain (the exact representation) $$\begin{align}\int_0^1\!dy \,e^{ix y^2}\cosh(y^2) = &e^{i\pi/4} \int_0^\infty\!dt\, e^{- x t^2}\cos(t^2) \\&- e^{i x} \int_0^\infty\!d\xi\,e^{-x\sinh(2\xi)} \cosh[1+i \sinh(2\xi) ] (\sinh\xi + i \cosh\xi).\end{align}$$

The expansion for $x\to\infty$ can now easily obtained by observing that in this limit the integrals are dominated for $t,\xi$ small. We have $$e^{i\pi/4} \int_0^\infty\!dt\, e^{- x t^2}\underbrace{\cos(t^2)}_{1-t^4/4+O(t^8)} =\frac{\sqrt\pi e^{i\pi/4}}{2} \left[ \frac1{x^{1/2}} - \frac{3}{8x^{5/2}} +O(x^{-9/2})\right]$$ and similarly $$\int_0^\infty\!d\xi\,e^{-x\overbrace{\sinh(2\xi)}^{2\xi +O(\xi^3)}} \underbrace{\cosh[1+i \sinh(2\xi) ] (\sinh\xi + i \cosh\xi)}_{i \cosh 1+ (\cosh 1 -2 \sinh 1) \xi+ O(\xi^2)} = \frac{i \cosh 1}{2x} + \frac{\cosh 1 -2 \sinh 1}{4 x^2} +O(x^{-3}) .$$

Thus, we have the result $$\int_0^1\!dy \,e^{ix y^2}\cosh(y^2) = \frac{\sqrt\pi e^{i\pi/4}}{2 x^{1/2}} -\frac{i e^{ix} \cosh 1}{2x} -\frac{e^{i x}(\cosh 1 -2 \sinh 1)}{4 x^2} - \frac{3\sqrt\pi e^{i\pi/4}}{16 x^{5/2}} + O(x^{-3})$$