Integration: Branch cuts

Question

Can someone show me how to calculate this integral using branch cuts ?

$$\int_0^{\infty}\Big(\frac{x}{1-x}\Big)^{\frac{1}{3}}\frac{1}{1+x^2}dx$$

Answer 1

Let $$f(z)=\frac{|z|^{1/3}e^{i\varphi_1/3}|z-1|^{-1/3}e^{-i\varphi_2/3}}{1+z^2}$$ where $\varphi_1=\arg{z}$, $\varphi_2=\arg(z-1)$, $0\le\varphi_1,\varphi_2\le2\pi$. Integrating $f(z)$ over a dumbbell contour, we get \begin{align} \int_\gamma f(z)\ dz &=\int^1_0\frac{x^{1/3}e^{i(0)/3}(1-x)^{-1/3}e^{-i(\pi)/3}}{1+x^2}dx+\int^0_1\frac{x^{1/3}e^{i(2\pi)/3}(1-x)^{-1/3}e^{-i(\pi)/3}}{1+x^2}dx\\ &=-i\sqrt{3}\int^1_0\frac{x^{1/3}(1-x)^{-1/3}}{1+x^2}dx\\ &=2\pi i\left(\sum_\pm\operatorname*{Res}_{z=\pm i}f(z)-\color{grey}{\operatorname*{Res}_{z=0}\frac{(1-z)^{-1/3}}{1+z^2}}\right)\\ &=2\pi i\left(\frac{e^{\pi i/6}\times2^{-1/6}\times e^{-\pi i/4}-e^{\pi i/2}\times2^{-1/6}\times e^{-5\pi i/12}}{2i}\right)\\ &=-i\frac{\pi}{\sqrt[3]{4}}\left(\sqrt{3}-1\right) \end{align} which implies $$\int^1_0\frac{x^{1/3}(1-x)^{-1/3}}{1+x^2}=\frac{\pi}{\sqrt[3]{4}}\left(1-\frac{1}{\sqrt{3}}\right)$$ For the remaining integral, one can apply the substitution $x\mapsto 1-x^3$ to get \begin{align} \int^1_0\frac{(x-1)^{-1/3}}{1+x^2} &=-\int^1_0\frac{3x}{(x^3-1)^2+1}dx\\ &=-\sum_{\omega\in S}\frac{1}{2(\omega^4-\omega)}\int^1_0\frac{x-\sigma+it}{(x-\sigma)^2+t^2}dx\\ &=-\sum_{\omega\in S}\frac{1}{2(\omega^4-\omega)}\left[\frac{1}{2}\ln\left((x-\sigma)^2+t^2\right)+i\arctan\left(\frac{x-\sigma}{t}\right)\right]^1_0\\ &=-\sum_{\omega\in S}\frac{\ln\left(\frac{(1-\sigma)^2+t^2}{\sigma^2+t^2}\right)+2i\left(\arctan\left(\frac{1-\sigma}{t}\right)+\arctan\left(\frac{\sigma}{t}\right)\right)}{4(\omega^4-\omega)} \end{align} where $\omega=\sigma+it$, $\sigma\in\mathbb{R}$ and $$S=\left\{\sqrt[6]{2}e^{\pi i/12},\sqrt[6]{2}e^{7\pi i/12},\sqrt[6]{2}e^{3\pi i/4},\sqrt[6]{2}e^{5\pi i/4},\sqrt[6]{2}e^{17\pi i/12},\sqrt[6]{2}e^{23\pi i/12}\right\}$$ Writing out the $6$ terms manually then using Mathematica to simplify, we find that the sum (which is equivalent to a hypergeometric function as pointed out by Lucian) evaluates to \begin{align} \frac{3}{2}{}_3\mathrm{F}_2\left(\left.\begin{matrix}1,1,\frac{1}{2}\\\frac{4}{3},\frac{5}{6}\end{matrix}\right|-1\right) =&\ \frac{\pi\sqrt[3]{2}}{12}(3+\sqrt{3})+\frac{\ln{2}}{3\sqrt[3]{4}}+\frac{\ln(\sqrt[3]{2}-1)}{4\sqrt[3]{4}}-\frac{\ln(2+\sqrt[3]{4}+\sqrt[3]{16})}{2\sqrt[3]{4}}\\ &+\frac{\sqrt{3}}{2\sqrt[3]{4}}\operatorname{artanh}\left(\sqrt{\frac{3}{5-\sqrt[3]{4}}}\right)+\frac{\sqrt{3}-1}{2\sqrt[3]{4}}\arctan\left(\frac{\sqrt{3}+\sqrt[3]{16}-1}{\sqrt{3}+1}\right)\\ &-\frac{\sqrt{3}+1}{2\sqrt[3]{4}}\arctan\left(\frac{\sqrt{3}-\sqrt[3]{16}+1}{\sqrt{3}-1}\right)-\frac{\arctan(1+\sqrt[3]{2})}{\sqrt[3]{4}} \end{align} Therefore, \begin{align} \int^\infty_0\frac{x^{1/3}(1-x)^{-1/3}}{1+x^2}dx =&\ -\frac{\pi\sqrt[3]{2}}{12}(3+\sqrt{3})-\frac{\ln{2}}{3\sqrt[3]{4}}-\frac{\ln(\sqrt[3]{2}-1)}{4\sqrt[3]{4}}+\frac{\ln(2+\sqrt[3]{4}+\sqrt[3]{16})}{2\sqrt[3]{4}}\\ &-\frac{\sqrt{3}}{2\sqrt[3]{4}}\operatorname{artanh}\left(\sqrt{\frac{3}{5-\sqrt[3]{4}}}\right)-\frac{\sqrt{3}-1}{2\sqrt[3]{4}}\arctan\left(\frac{\sqrt{3}+\sqrt[3]{16}-1}{\sqrt{3}+1}\right)\\ &+\frac{\sqrt{3}+1}{2\sqrt[3]{4}}\arctan\left(\frac{\sqrt{3}-\sqrt[3]{16}+1}{\sqrt{3}-1}\right)+\frac{\arctan(1+\sqrt[3]{2})}{\sqrt[3]{4}}\\ &+\frac{\pi}{\sqrt[3]{4}}\left(1-\frac{1}{\sqrt{3}}\right) \end{align}

Answer 2

Using the simple substitution $x\mapsto \tan(x)$, we get that

$$\int_0^{\infty}\Big(\frac{x}{1-x}\Big)^{\frac{1}{3}}\frac{1}{1+x^2}dx=\int_0^{\pi/2}\Big(\frac{\tan(x)}{1-\tan(x)}\Big)^{\frac{1}{3}} \ dx$$

where the last integral can be computed by Mathematica as shown below.

$$\frac{1}{48 \sqrt[3]{2}}\Re\left(\sqrt{3} \log (4) \sqrt[3]{-1-i}-24 i \log \left(-\sqrt[3]{-\frac{1}{2}-\frac{i}{2}}\right) \sqrt[3]{-1-i}+12 i \log \left(-(-1)^{2/3} \sqrt[3]{-\frac{1}{2}-\frac{i}{2}}\right) \sqrt[3]{-1-i}+12 \sqrt{3} \log \left(-(-1)^{2/3} \sqrt[3]{-\frac{1}{2}-\frac{i}{2}}\right) \sqrt[3]{-1-i}+24 i \log \left(\sqrt[3]{-2}-\sqrt[3]{-1-i}\right) \sqrt[3]{-1-i}-12 i \log \left(\sqrt[3]{-2}-(-1)^{2/3} \sqrt[3]{-1-i}\right) \sqrt[3]{-1-i}-12 \sqrt{3} \log \left(\sqrt[3]{-2}-(-1)^{2/3} \sqrt[3]{-1-i}\right) \sqrt[3]{-1-i}-12 i \log \left(\sqrt[3]{1+i}+\sqrt[3]{-2}\right) \sqrt[3]{-1-i}+12 \sqrt{3} \log \left(\sqrt[3]{1+i}+\sqrt[3]{-2}\right) \sqrt[3]{-1-i}+\left(-1-i \sqrt{3}\right) \pi \sqrt[3]{-1-i}+\left(-2 i \sqrt[3]{-1-i}-8 i \sqrt[3]{-1+i}+8 \sqrt{3} \sqrt[3]{-1+i}\right) \log (2)-12 i \sqrt[3]{-1+i} \log \left(\sqrt[3]{-1+i} \sqrt[3]{-\frac{1}{2}}\right)+12 \sqrt{3} \sqrt[3]{-1+i} \log \left(\sqrt[3]{-1+i} \sqrt[3]{-\frac{1}{2}}\right)+24 \sqrt[3]{-1+i} i \log \left(-\sqrt[3]{-\frac{1}{2}+\frac{i}{2}}\right)-12 i \sqrt[3]{-1+i} \log \left(-(-1)^{2/3} \sqrt[3]{-\frac{1}{2}+\frac{i}{2}}\right)-12 \sqrt[3]{-1+i} \sqrt{3} \log \left(-(-1)^{2/3} \sqrt[3]{-\frac{1}{2}+\frac{i}{2}}\right)+12 \sqrt[3]{-1+i} i \log \left(\sqrt[3]{-1} \left(2^{2/3} \sqrt[3]{-1+i}+2\right)\right)-12 \sqrt[3]{-1+i} \sqrt{3} \log \left(\sqrt[3]{-1} \left(2^{2/3} \sqrt[3]{-1+i}+2\right)\right)-24 i \sqrt[3]{-1+i} \log \left(\sqrt[3]{-2}-\sqrt[3]{-1+i}\right)+12 \sqrt{3} \sqrt[3]{-1+i} \log \left(\sqrt[3]{-2}-(-1)^{2/3} \sqrt[3]{-1+i}\right)+12 \sqrt[3]{-1+i} i \log \left(\sqrt[3]{-2}-(-1)^{2/3} \sqrt[3]{-1+i}\right)\right)$$