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

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 $0$?

• bender describes also the method using ibp in his book – tired Nov 18 '16 at 10:11
• @tired I'm not familiar with the book by Bender, but pursued this problem yesterday and posted a solution. Does this approach look similar to Bender's? – Mark Viola Nov 18 '16 at 19:15

## 2 Answers

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!

• @tired I figured out the "issue" after all. The IBP scheme in the previous development rendered the remainder term of order $1/x$ always. So, the expansion failed. I split the integral in this newer development and things work out perfectly fine! – Mark Viola Nov 18 '16 at 21:36
• now it is fine (+1). That's more or less what Bender suggests if one wants to approximate such type of integrals using IPBs. Btw you should somehow get a hand on this book it is one of the best textbooks i ever get in touch with. I can also recommend the youtube lecturers by Prof. Bender. They are also great stuff – tired Nov 20 '16 at 15:16
• @tired Thank you! And Bender's book sounds great. What is the title? – Mark Viola Nov 20 '16 at 18:00
• advanced mathematical methods for scientists and engineers – tired Nov 20 '16 at 18:03
• How does it compare with the books by Morse and Feshback or Courant and Hilbert? – Mark Viola Nov 20 '16 at 18:25

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.

• The OP wants to develop the large $x$ asymptotic expansion using IBP. You have not provided a large $x$ expansion. – Mark Viola Nov 18 '16 at 19:26
• WRONG!!!!!!!!!! – tired Nov 20 '16 at 19:06