Pizza-slice contour

I occasionally browsed a youtube video on "2023 MIT Integration Bee". I thought it is a competition between students taking the course of calculus. I took a try and it turns out that I was not able to even solve the first problem in the allowed 4 mins. 

The first problem is to solve \begin{equation} I \equiv \int_0^{\infty} \frac{\sqrt[3]{\tan x}}{(\sin x + \cos x)^2}dx\,.\tag{1}\end{equation} It is obvious that \begin{eqnarray}I &=&  \int_0^{\infty} \frac{\sqrt[3]{\tan x}}{(\tan x + 1)^2} \frac{dx}{\cos^2 x} =  \int_0^{\infty} \frac{\sqrt[3]{x}}{(x+1)^2}dx\\&=&-\left. \frac{\sqrt[3]{x}}{x+1}\right|_0^{\infty} +  \int_0^{\infty} \frac{d\,\sqrt[3]{x}}{x+1}\\&=&  \int_0^{\infty}\frac{dx}{x^3+1}\,.\tag{2}\end{eqnarray}
I know how to solve the indefinite integral $\int\frac{dx}{x^3+1}$, but I was not able to accomplish the entire calculation within 4 mins. Here is another youtube video that provides step-by-step calculations. The result is \begin{equation}\int\frac{dx}{x^3+1}=\frac{1}{6} \log \frac{(x+1)^2}{x^2-x+1} + \frac{1}{\sqrt{3}}\,\arctan \left(\frac{2x-1}{\sqrt{3}}\right)\,.\end{equation} As a result, we have \begin{equation} \int_0^{\infty}\frac{dx}{x^3+1} = \frac{1}{\sqrt{3}}\left(\frac{\pi}{2} - \left(-\frac{\pi}{6}\right) \right)=\frac{2\sqrt{3}}{9}\pi\,.\end{equation}

There is another way to calculate the definite integral $\int_0^{\infty}\frac{dx}{x^3+1}$ without doing the indefinite integral $\int\frac{dx}{x^3+1}$. Consider the following pizza-slice contour in the complex plane:

Fig. 1. The pizza-slice contour for the integral (2)

For such contour, we know that \begin{eqnarray} \oint_C \frac{dz}{z^3+1} &=& 2\pi i\, \text{Res}\left[\frac{1}{z^3+1}, e^{i\pi /3}\right]\\ &=& 2\pi i\,\lim_{z\rightarrow e^{i\pi /3}} \frac{z-e^{i\pi /3}}{z^3+1}=2\pi i\,\lim_{z\rightarrow e^{i\pi /3}}\frac{1}{3z^2} = \frac{2\pi i}{3 e^{i2\pi /3}} \,.\tag{3}\end{eqnarray} On the other hand, $\oint_C \frac{dz}{z^3+1}=\int_{C_1} \frac{dz}{z^3+1} +\int_{C_2} \frac{dz}{z^3+1}  + \int_{C_3} \frac{dz}{z^3+1}$. The first term is our integral $I$ as in Eq. (2). The second term is \begin{equation} \int_{C_2} \frac{dz}{z^3+1} =\lim_{R\rightarrow\infty} \int_{0}^{2\pi/3} \frac{R}{R^3e^{i3\theta}+1}de^{i\theta} = 0\,,\end{equation} while the third term is \begin{equation} \int_{C_3} \frac{dz}{z^3+1} = - \int_{0}^{\infty} \frac{e^{i2\pi/3}}{r^3+1}dr=-e^{i2\pi/3} I\,.\end{equation} Therefore, we have \begin{equation} I - e^{i2\pi/3} I + 0 =  \frac{2\pi i}{3 e^{i2\pi /3}}\,,\tag{4}\end{equation} from which we can solve \begin{equation} I =\frac{2\pi i}{3 e^{i2\pi /3}(1-e^{i2\pi/3})}=\frac{2\sqrt{3}}{9}\pi\,.\end{equation} In general, following the above recipe, we can calculate \begin{equation} \int_0^{\infty}\frac{dx}{x^n+1} = \frac{2\pi i}{n\,e^{i(n-1)\pi/n}\,(1-e^{i2\pi/n})}=\frac{\pi}{n\sin \frac{\pi}{n}}\,.\tag{5} \end{equation}

BTW, in my impression, the first time I saw the application of pizza-slice contour in physics is in the undergraduate quantum mechanics course. The Green function of free particle in quantum mechanics is \begin{equation} G(x,t) = \int \frac{d^3p}{(2\pi \hbar)^3} \exp^{i p\cdot x /\hbar -i p^2t/(2m\hbar)}\,,\end{equation} which can be reduced to the calculation of the integral \begin{equation} \int_0^{\infty} dx\, e^{ix^2}\tag{6}\end{equation} after some algebra. To solve the integral (6), we can natively let $a=-i$ in the Gaussian integral: \begin{equation} \int_0^{\infty} dx\, e^{-ax^2}=\frac{1}{2} \int_{-\infty}^{\infty} dx\, e^{-ax^2}=\frac{1}{2}\sqrt{\frac{\pi}{a}}\,.\end{equation} To justify the correctness of this naive approach, we consider another pizza-slice contour as in Fig. 2: 

Fig. 2. The pizza-slice contour for the integral (6)

 \begin{equation}  \int_{C_1} dz e^{iz^2} +  \int_{C_2} dz e^{iz^2}+ \int_{C_3} dz e^{iz^2} = \oint_C dz e^{iz^2} = 0\,. \end{equation} The first term is the integral (6). The second term is 0 as $R\rightarrow \infty$. The third term is \begin{equation}\int_{C_3} dz e^{iz^2} = -\int_{0}^{\infty} e^{i\pi/4} dr \,e^{i(re^{i\pi/4})^2}=-e^{i\pi/4} \int_{0}^{\infty} dr e^{-r^2} = -\frac{\sqrt{\pi}}{2}e^{i\pi/4}=-\frac{1}{2}\sqrt{i\pi}\,.\end{equation} 

The intuition of constructing the pizza-slice contour in Fig. 2 is simply to "rotate" the integral along the real axis to another line $re^{-i\theta}$ for some $\theta$ such that the integrand along the new line becomes Gaussian $e^{-r^2}$.  With exactly the same motivation, in quantum field theory, a more commonly rotations is the so-called Wick rotation. The corresponding double pizza-slice contour shown as in Fig. 3, which transforms the integration along the real axis to the imaginary axis. 

Fig 3.  The contour behind the Wick rotation. This picture is copied from Prof. Schwarz's QFT textbook.