Cauchy's Integral with quadratic exponential term

As I was studying the Cauchy's integral formula, I tried to do the integral:

$$$$I = \int\limits_{-\infty}^{\infty} \frac{1}{x - a} e^{(i A x^2 + i B x)} dx$$$$ with $$A>0, B>0$$ and $$a > 0$$.

Consider an integral on a complex plan: $$$$J = \int\limits_{C + C_R} \frac{1}{z - a} e^{(i A z^2 + i B z)} dz$$$$ where $$C$$ is along the real axis $$-\infty \rightarrow +\infty$$ and $$C_R$$ is the upper half circle $$z = Re^{i\theta}$$ with $$R \rightarrow \infty$$ and $$\theta \in [0, \pi]$$.

Naively, I would expect $$C_R$$ part of the integral gives zero and $$C$$ part of the integral gives $$I$$, then the $$I$$ can be derived by Cauchy's integral formula.

However, as I tried to check the $$C_R$$ part of the integral, I found that ($$z = Re^{i\theta}$$): $$\begin{split} I_R &= \int\limits_0^{\pi} d\theta \frac{iRe^{i\theta}}{Re^{i\theta} - a} \exp\big(iAR^2e^{2i\theta}+iBRe^{i\theta}\big) \\ |I_R| &\leq \int\limits_0^{\pi} d\theta\left |\frac{iRe^{i\theta}}{Re^{i\theta} - a}\right| \Big|\exp\big(iAR^2e^{2i\theta}+iBRe^{i\theta}\big)\Big| \end{split}$$ where the first term

$$$$\left|\frac{iRe^{i\theta}}{Re^{i\theta} - a}\right| \leq \frac{R}{R-a} \rightarrow 1 \ as\ R \rightarrow \infty$$$$

and the second term $$$$\left|\exp(iAR^2e^{2i\theta}+iBRe^{i\theta})\right| \leq e^{-AR^2\sin(2\theta) - BR\sin(\theta)}$$$$ will not approach to zero because of $$e^{-AR^2\sin(2\theta)}$$.

Is there anything wrong in my approach? And is there any other way I can perform this integral $$I$$?

Thanks a million for advises!

• Have you tried a rectangle instead of circle? – DuFong Jun 1 at 6:32
• @DuFong. Thanks for your suggestion. Yes, I tried a contour $C_1:-R \rightarrow R$, $C_2: R \rightarrow R+i\epsilon$, $C_3: R+i\epsilon \rightarrow -R+i\epsilon$ and $C_4: -R+i\epsilon \rightarrow -R$. The integrals of $C_2$ and $C_4$ approach to zero as $R\rightarrow \infty$. But the one of $C_3$ faces the same problem : $\int_{-\infty}^{\infty} dx e^{-2A\ x\ \epsilon}$ will not. Any other suggestions, please? – physics_rocks Jun 1 at 9:43
• Before anything else, you should define what you mean by integrating over the point $x=a$. Is this meant to be a principal value? – Michael Engelhardt Jun 1 at 13:03
• This integral is apparently related to the time-dependent linear Schrodinger equation with step function initial data, have you check the StackExchange: this. And to study this, inverse scattering is possible the most commonly used method in mathematical physics. By the way, in order to make sense, $\frac{1}{x-a}$ should be regarded as a distribution. As it is not easy to find the exact solution, one may consider finding asymptotics as $A\rightarrow \infty$ – DuFong Jun 1 at 13:57
• @MichaelEngelhardt. Yes, it's a principal value. – physics_rocks Jun 2 at 7:27

1 Answer

Let me first remove the $$Bx$$ term by completing the square, $$I=\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2+iBx}}{x - a}\,dx=e^{-iB^2/4A}\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2}}{x - a-B/2A}\,dx.$$ Mathematica evaluates the Cauchy principal value of the integral in terms of Meijer G-functions, $$I=-\tfrac{1}{8} \pi ^{-5/2} e^{-iB^2/4A}\biggl\{G_{3,5}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} 0,\frac{1}{4},\frac{3}{4} \\ 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)+8 \pi ^4 G_{7,9}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} 0,\frac{1}{4},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ \end{array} \right)+i G_{3,5}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} \frac{1}{4},\frac{1}{2},\frac{3}{4} \\ 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)+8 \pi ^4 i G_{7,9}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} \frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ \end{array} \right)\biggr\},$$ with $$\alpha=\left(a+\frac{B}{2A}\right)^4\frac{A^2}{4}.$$