I need to find the first three terms in the asymptotic expansion of given integral $$\int_0^1ln(1+t)e^{ixsin^2t}dt$$ as $x$ tends to infinity.
Using steepest descent method, we deform contour C:$0\lt t \le 1$ into contours along which $\Im A(t)$ is constant, where $A(t)= i \sin^2t$.
To find constant contour along $t=0$, set $\sin t=u+iv$, $i \sin^2t=-2uv+i(u^2-v^2)$. Thus $\Im A(t)=u^2-v^2$.
At $t=0$, $\Im A(t) =0$. Thus, at $t=0$, $u=v$. $\Re A(t)=-2v^2$. Hence, $e^{ix \sin^2t}=e^{-2xv^2}$.
I think I'm in the wrong way, can anybody help me to solve this.
Here's the general idea of the contour deformation part of applying the method of steepest descent to your problem.
Let $g(t) = i(\sin t)^2$. Note that $\DeclareMathOperator{\re}{Re} \re g(t) = 0$. We will deform the integration contour so that, except its endpoints at $t=0$ and $t=1$, it lies completely in the region $\re g(t) < 0$. Near its endpoints we will take the contour to lie tangent to the paths of steepest descent from the endpoints.
Near the endpoint $t=0$ we have
$$ g(t) = it^2 + O(t^4), $$
so the steepest descent path on the surface $\re g(t)$ through the point $t=0$ leaves this point at an angle of $\pi/4$.
Near the endpoint $t=1$ we have
$$ g(t) = i(\sin 1)^2 + i(\sin 2)(t-1) + O\!\left((t-1)^2\right), $$
so the steepest descent path through the point $t=1$ leaves this point at an angle of $\pi/2$.
So, we deform the interval $[0,1]$ into a curve which leaves the point $t=0$ at an angle of $\pi/4$ and returns to the point $t=1$ at an angle of $-\pi/2$.
Here is a plot of the original contour $[0,1]$ in black and the new contour in $\color{red}{\textbf{red}}$. In the background several level curves of the surface $\re g(t)$ are shown, with higher parts of the surface in orange and lower parts in blue.
Now, as $x \to \infty$, the main contributions to the integral come from neighborhoods of the points $t=0$ and $t=1$, and near each of these points the integral can be approximated using the Laplace method.