Suppose \(f\colon\RR\to\CC\) satisfies \(f'(t)+2itf(t)=e^{2it}\) and \(f(0)=0\). Compute
\[ \lim_{t\to+\infty}e^{it^2}(f(t)-f(-t)). \]
You may assume \(\int_0^\infty e^{-t^2}\,dt=\sqrt\pi/2\).
Solution
Multiply the ODE by the integrating factor \(e^{it^2}\), and integrate to get
\[ e^{it^2}f(t)=\int_0^te^{ix^2+2ix}\,dx \]
(The hypothesis \(f(0)=0\) implies that there is no constant of integration.) Substituting \(-t\) for \(t\) and subtracting, we get
\[ \begin{aligned} e^{it^2}(f(t)-f(-t))&=\int_{-t}^te^{ix^2+2ix}\,dx\\ &=e^{-i}\int_{-t}^te^{i(x+1)^2}\,dx\\ &=e^{-i}\int_{-t+1}^{t+1}e^{iz^2}\,dz. \end{aligned} \]
Since \(e^{iz^2}\) is an even function, the limit as \(t\to+\infty\) equals \(2e^{-i}I\), where \(I\coloneqq\lim_{R\to+\infty}\int_0^Re^{iz^2}\,dz\) (assuming for now that the latter limit exists). Apply Cauchy’s theorem to the triangular contour from \(0\) to \(R\) to \(R+Ri\) and back to \(0\). The vertical part contributes
\[ \int_R^{R+Ri}e^{iz^2}\,dz=\int_0^Re^{i(R+ti)^2}\,i\,dt, \]
whose absolute value is bounded by
\[ \begin{aligned} \int_0^R\abs{e^{i(R+ti)^2}}\,dt&=\int_0^Re^{-2Rt}\,dt\\ &=\frac{1}{2R}\int_0^{2R^2}e^{-u}\,du, \end{aligned} \]
which goes to \(0\) as \(R\to\infty\). Thus
\[ \begin{aligned} I&=\lim_{R\to\infty}\int_0^{R+Ri}e^{iz^2}\,dz&&\text{(if the limit exists)}\\ &=\lim_{R\to\infty}\int_0^Re^{i(e^{i\pi/4}t)^2}e^{i\pi/4}\,dt&&\text{(if the limit exists)}\\ &=e^{i\pi/4}\lim_{R\to\infty}\int_0^Re^{-t^2}\,dt&&\text{(if the limit exists)}\\ &=e^{i\pi/4}\frac{\sqrt\pi}{2}. \end{aligned} \]
Thus we now know that all the limits exist, and the answer is \(2e^{-i}I=e^{-i+i\pi/4}\sqrt\pi\).