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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.

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  • $\begingroup$ your subsitution is strange to say the least... $\endgroup$
    – tired
    Mar 23, 2017 at 22:13
  • $\begingroup$ @tired, I'm quite weak in this. Can you please help me to do this? $\endgroup$
    – user420487
    Mar 23, 2017 at 22:51

1 Answer 1

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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 $-\pi/2$.

contour deformation

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.

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  • $\begingroup$ A good reference for this is section 5.8 of de Bruijn's Asymptotic Methods in Analysis. $\endgroup$ Mar 30, 2017 at 23:50

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