Return to Answer

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

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 $$\alpha=\left(a+\frac{B}{2A}\right)^4\frac{A^2}{4}.$$

For what it's worth, the Cauchy principal value of the integral for $B=0$ has a lengthy expression in terms of Meijer G-functions, $$I=-8 \pi ^{5/2} \int\limits_{-\infty}^{\infty} \frac{e^{i A x^2}}{x - a} \, dx=G_{3,5}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} 0,\frac{1}{4},\frac{3}{4} \\ 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)+8 \pi ^4 G_{7,9}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} 0,\frac{1}{4},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ \end{array} \right)+i G_{3,5}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} \frac{1}{4},\frac{1}{2},\frac{3}{4} \\ 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)+8 \pi ^4 i G_{7,9}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} \frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ \end{array} \right),$$ with $\alpha=a^4A^2/4$. The large-$\alpha$ asymptotics is $$I\rightarrow i \pi e^{ 2i\sqrt\alpha}-\frac{(1+i) \sqrt\pi}{2\alpha^{1/4}}.$$

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

For what it's worth, the Cauchy principal value of the integral for $B=0$ has a lengthy expression in terms of Meijer G-functions, $$-8 \pi ^{5/2} \int\limits_{-\infty}^{\infty} \frac{e^{i A x^2}}{x - a} \, dx=G_{3,5}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} 0,\frac{1}{4},\frac{3}{4} \\ 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)+8 \pi ^4 G_{7,9}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} 0,\frac{1}{4},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ \end{array} \right)+i G_{3,5}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} \frac{1}{4},\frac{1}{2},\frac{3}{4} \\ 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)+8 \pi ^4 i G_{7,9}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} \frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ \end{array} \right),$$ with $\alpha=a^4A^2/4$.

For what it's worth, the Cauchy principal value of the integral for $B=0$ has a lengthy expression in terms of Meijer G-functions, $$I=-8 \pi ^{5/2} \int\limits_{-\infty}^{\infty} \frac{e^{i A x^2}}{x - a} \, dx=G_{3,5}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} 0,\frac{1}{4},\frac{3}{4} \\ 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)+8 \pi ^4 G_{7,9}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} 0,\frac{1}{4},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ 0,0,\frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ \end{array} \right)+i G_{3,5}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} \frac{1}{4},\frac{1}{2},\frac{3}{4} \\ 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4} \\ \end{array} \right)+8 \pi ^4 i G_{7,9}^{5,3}\left(\alpha\,\biggl| \begin{array}{c} \frac{1}{4},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ 0,\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4},-\frac{1}{8},\frac{1}{8},\frac{3}{8},\frac{5}{8} \\ \end{array} \right),$$ with $\alpha=a^4A^2/4$. The large-$\alpha$ asymptotics is $$I\rightarrow i \pi e^{ 2i\sqrt\alpha}-\frac{(1+i) \sqrt\pi}{2\alpha^{1/4}}.$$