how to find the full asymptotic expansion of this integral by integration by parts?

Question

Consider $\int_{0}^{1}e^{ixt^2}dt$. It is easy to find the leading order of this by change of variable $t^2=s$, scaling, and apply some change of contour integration. And we can find the full asymptotic expansion using steepest descent. In Bender's book, before he did it using steepest descent, he mentioned that we can use integration by parts. But how can we apply integration by parts given that we can not evaluate the integral we separated near $0$?

Answer 1

Let $I(x)$ be the integral given by

$$\bbox[5px,border:2px solid #C0A000]{I(x)=\int_0^1e^{ixt^2}\,dt} \tag 1$$

Note that we can rewrite $(1)$ as

$$\begin{align} I(x)&=\int_0^\infty e^{ixt^2}\,dt-\int_1^\infty e^{ixt^2}\,dt\\\\ &=\frac12 \sqrt{\frac{\pi}{x}}e^{i\pi/4}-\int_1^\infty e^{ixt^2}\,dt \tag2 \end{align}$$

We now proceed to use integration by parts to develop the first order term in the large $x$ asymptotic expansion for $I(x)$.

---

Integrating by parts the integral in $(2)$ with $u=\frac1t$ and $v=\frac{e^{ixt^2}}{i2x}$ reveals

$$\begin{align} I(x)&=\frac12 \sqrt{\frac{\pi}{x}}e^{i\pi/4}-\left.\left(\frac1t \left(\frac{e^{ixt^2}}{i2x}\right)\right)\right|_{1}^{\infty}+\int_1^\infty \left(\frac{e^{ixt^2}}{i2xt^2}\right)\,dt\\\\ &=\frac12 \sqrt{\frac{\pi}{x}}e^{i\pi/4}+\frac{e^{ix}}{i2x}+\int_1^\infty \left(\frac{e^{ixt^2}}{i2xt^2}\right)\,dt\tag 3 \end{align}$$

---

The integral on the right-hand side of $(3)$ is of order $1/x^2$. To see this, we integrate by parts again with $u=\frac1{t^3}$ and $v=\frac{e^{ixt^2}}{(i2x)^2}$ and find that

$$\begin{align} \int_1^\infty \left(\frac{e^{ixt^2}}{i2xt^2}\right)\,dt&=\left.\left(\frac1{t^3}\frac{e^{ixt^2}}{(i2x)^2}\right)\right|_{1}^{\infty}-3\int_1^\infty \frac{e^{ixt^2}}{(i2x)^2\,t^4}\,dt\\\\ &=-\frac14\frac{e^{ix}}{x^2}-3\int_1^\infty \frac{e^{ixt^2}}{(i2x)^2\,t^4}\,dt\\\\ &=O\left(\frac1{x^2}\right)\tag 4 \end{align}$$

By induction, the integral on the right-hand side of $(4)$ is of order $1/x^3$.

---

Therefore, the second order asymptotic expansion for large $x$ of $(1)$ is

$$\bbox[5px,border:2px solid #C0A000]{I(x)\sim \frac12 \sqrt{\frac{\pi}{x}}e^{i\pi/4}+\frac12\left(\frac{e^{ix}}{ix}\right)+\frac14 \left(\frac{e^{ix}}{(ix)^2}\right)}$$

as was to be shown using integration by parts!

Answer 2

If we assign $u$ and $v$ as such:

$$\int_0^1\underbrace{1}_{u'}\cdot \underbrace{e^{ixt^2}}_v\,dt\Rightarrow \left\{\begin{array}{l} u=t\\v'= 2ixt\cdot e^{ixt^2}\end{array}\right.$$

Hence, integration by parts yields:

\begin{align} \int_0^1e^{ixt^2}\,dt&={\left[t e^{ixt^2}\right]}_0^1\,\,-\int_0^12ixt^2\cdot e^{ixt^2}\,dt\\ &=e^{ix}-2ix\int_0^1t^2e^{ixt^2}\,dt \end{align}

I don't exactly see how there's any problem with evaluation near $0$ anywhere.