Finding the improper integral

Question

I was asked a question, to evaluate the improper integral: $$\int \limits_{0}^{\infty}\frac{x^{1/3}}{1+x^2}dx$$

I am using complex analysis to solve this. Consider some small radius r, large radius R, and small angle η. I am considering the integral along γ1 + γR + γ2 + γr, where γR is the circle of radius R, omitting points with argument between −η and η, γr is clockwise around the circle of radius r, omitting points with argument between −η and η, γ1 is along the ray of argument η, from γr to γR, and γ2 is along the ray of argument −η, from γR to γr. I think the integral along γR and γr is 0. I estimated them by ML inequality and the integral along those tend to 0. But I am not able to evaluate the integral along γ1 and γ2. I think I can employ the residue theorem for them. But I do not know what the index is. Any help is appreciated.

Answer 1

Consider the contour integral \begin{align} \int_C f(z)\ dz=\int_{C}\frac{z^{1/3}}{1+z^2}\ dz \end{align} where $C=L_1+C_R +L_2 +C_\epsilon$ is given by

where \begin{align} z^{1/3} = \exp\left(\frac{1}{3}\log z \right) \end{align} is given by the branch of the logarithm with $-\frac{\pi}{2}<\theta \le \frac{3\pi}{2}$.

It's not hard to see that $f(z)$ has a simple pole at $z=i$ and analytic everywhere else on the strictly upper half plane. Hence by Cauchy's theorem, we have that \begin{align} \int_{L_1} \frac{z^{1/3}}{1+z^2}\ dz + \int_{C_R}\frac{z^{1/3}}{1+z^2}\ dz+ \int_{L_2}\frac{z^{1/3}}{1+z^2}\ dz+\int_{C_\epsilon}\frac{z^{1/3}}{1+z^2}\ dz = 2\pi i \operatorname{Res}_{z= i} f(z) = \pi i^{1/3} = \pi e^{i\pi/6}. \end{align} Let us simplify each integral. Observe \begin{align} \int_{L_1} \frac{z^{1/3}}{1+z^2}\ dz = \int^R_{\epsilon} \frac{x^{1/3}}{1+x^2}\ dx \end{align} and \begin{align} \int_{L_2} \frac{z^{1/3}}{1+z^2}\ dz =&\ \int^{-\epsilon}_{-R} \frac{x^{1/3}}{1+x^2}\ dx . \end{align}

Next, observe that \begin{align} \left|\int_{C_R}\frac{\exp\left(\frac{1}{3}\log|z|+i\frac{\theta}{3} \right)}{1+z^2}\ dz\right|\le \int_{C_R} \frac{R^{1/3}}{R^2-1}\ |dz| \le C \frac{R^{1+1/3}}{R^2-1}\rightarrow 0 \end{align} as $R \rightarrow \infty$. Lastly, we have that \begin{align} \left|\int_{C_\epsilon} \frac{z^{1/3}}{1+z^2}\ dz\right|\le \frac{C\epsilon^{1/3+\epsilon}}{1-\epsilon^2}\Rightarrow 0 \end{align} as $\epsilon\rightarrow 0$. Thus, it follows \begin{align} \int^\infty_{-\infty} \frac{x^{1/3}}{1+x^2}\ dx = \pi e^{i\pi/6}. \end{align}

Edit: This post was made when the original question was to evaluate \begin{align} \int^\infty_{-\infty} \frac{x^{1/3}}{1+x^2}\ dx. \end{align} The reader should note that this integral doesn't have a unique answer. You could get what I have shown or you could also get what robjohn have gotten. The ambiguity comes from the interpretation of $x^{1/3}$, which is not as simple as it looks. In particular, we need to first determine the meaning of $(-1)^{1/3}$. Most students in high school or even freshmen in college are taught that $(-1)^{1/3} = -1$ since $(-1)^3 = -1$, which is correct since a first course in calculus is usually restricted to the real numbers.

However, once students have learned a bit about the complex numbers then they should start to re-evaluate what $(-1)^{1/3}$ actually means. The first thing they should do is to look at any pre-calculus books (in particular, they should look at the pre-calculus book they had learned from) to see the definition of $a^b$. It shouldn't come as a surprise that most of the books only define $a^b$ for $a>0$.

So how should we define $(-1)^{1/3}$? The general definition used in complex analysis for raising a complex number $z$ to a complex power $a$ is defined as follow

\begin{align} z^a = \exp\left(a \log z \right) = \exp\left( a\ast [\log|z|+i \arg \theta]\right), \end{align}

i.e. you need to make a choice for your range of $\theta$ (in other words, you need to choose a branch of the logarithm) otherwise $z^a$ will be a multi-valued function on the complex plane, which mean it is not a function.

Why is it not a function? Let us look at the example $(-1)^{1/3}$. By definition \begin{align} (-1)^{1/3} = \exp\left(\frac{1}{3}\log(-1) \right) = \exp\left(\frac{i}{3}\arg (-1) \right). \end{align} Let us test the cases $\theta = \pi, 3\pi, 5\pi$ (all the other cases are similar to these three) since all three are arguments of $-1$. Observe \begin{align} \exp\left(i\frac{\pi}{3} \right) =&\ \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}= \frac{1}{2}+i\frac{\sqrt{3}}{2}\\ \exp\left(i\frac{3\pi}{3} \right) =&\ \cos \pi + i \sin \pi = -1\\ \exp\left(i\frac{5\pi}{3} \right) =&\ \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} = \frac{1}{2}-i\frac{\sqrt{3}}{2} \end{align} which means $(-1)^{1/3}$ have three interpretations.

In my interpretation of $z^{1/3}$, I have used the branch where $-\frac{\pi}{2}<\theta\le \frac{3\pi}{2}$, which contains $\pi$.Whereas, robjohn chose a branch that contains $3\pi$ ( or something similar) say $\frac{3\pi}{2}<\theta\le\frac{7\pi}{2}$ which will lead to the conclusion \begin{align} \int^\infty_{-\infty} \frac{x^{1/3}}{1+x^2}\ dx = 0. \end{align}

Answer 2

Since $x^{1/3}$ is an odd function on $\mathbb{R}$, $$ \int_{-\infty}^\infty\frac{x^{1/3}}{1+x^2}\,\mathrm{d}x=0\tag{1} $$

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However, if we want to compute the integral over the positive real axis, then $$ \int_0^\infty\frac{x^{1/3}}{1+x^2}\,\mathrm{d}x =\frac12\int_0^\infty\frac{x^{-1/3}}{1+x}\,\mathrm{d}x\tag{2} $$ The contour integral $$ \begin{align} \int_\gamma\frac{z^{-1/3}}{1+z}\,\mathrm{d}z &=\left(1-e^{-2\pi i/3}\right)\int_0^\infty\frac{x^{-1/3}}{1+x}\,\mathrm{d}x\\ 2\pi ie^{-\pi i/3} &=e^{-\pi i/3}\left(e^{\pi i/3}-e^{-\pi i/3}\right)\int_0^\infty\frac{x^{-1/3}}{1+x}\,\mathrm{d}x\\ \pi&=\sin(\pi/3)\int_0^\infty\frac{x^{-1/3}}{1+x}\,\mathrm{d}x\\ \frac{2\pi}{\sqrt3}&=\int_0^\infty\frac{x^{-1/3}}{1+x}\,\mathrm{d}x\tag{3} \end{align} $$ where $\gamma$ is the curve

Combining $(2)$ and $(3)$, we get $$ \int_0^\infty\frac{x^{1/3}}{1+x^2}\,\mathrm{d}x=\frac\pi{\sqrt3}\tag{4} $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Also, you can still evaluate it by '$\ds{\color{#f00}{real\ methods}}$' as follows: $$ x \equiv \pars{{1 \over t} - 1}^{1/2}\iff t = {1 \over 1 + x^{2}}\,,\qquad \totald{x}{t} = -\,{t^{1/2}\pars{1 - t}^{-1/2} \over 2t^{2}} $$

\begin{align} \int_{0}^{\infty}{x^{1/3} \over 1 + x^{2}}\,\dd x & = \int_{0}^{1}t\bracks{\pars{1 - t}^{1/6}\,t^{-1/6}}\, {t^{1/2}\pars{1 - t}^{-1/2} \over 2t^{2}}\,\dd t = {1 \over 2}\int_{0}^{1}t^{-2/3}\pars{1 - t}^{-1/3}\,\dd t \\[5mm] & = {1 \over 2}\,\mrm{B}\pars{{1 \over 3},{2 \over 3}}\qquad \pars{~\mrm{B}:\ Beta\ Function~} \\[5mm] & = {1 \over 2}\,{\Gamma\pars{1/3}\Gamma\pars{2/3} \over \Gamma\pars{1/3 + 2/3}} \qquad\pars{~\Gamma:\ Gamma\ Function.\ \mbox{Note that}\ \Gamma\pars{1} = 1~} \\[5mm] & = {1 \over 2}\,{\pi \over \sin\pars{\pi/3}}\qquad \pars{~Euler\ Reflection\ Formula~} \\[5mm] & = {1 \over 2}\,{\pi \over \root{3}/2} = \bbx{\ds{{\root{3} \over 3}\,\pi}} \end{align}