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This question was asked in a masters of mathematics exam for which I am preparing.

Compute $\int\limits_0^{\infty} \frac{x^{1/3}}{1+x^{2}} dx$.

I could only think of substituting $y^3 = x$ and that does not change much.

Could somebody post a solution using residues or in ways besides here in this link?: How to compute the integral $\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$

Edit : I am interested in the answers which use contour integration and residue calculus.

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  • $\begingroup$ While not necessarily the easiest, the Calc II approach with $y=x^3$ gives $$ \int_0^{\infty}\frac{y}{1+y^6}3y^2dy $$ which can be integrated via partial fractions. $\endgroup$
    – yoyo
    Sep 2 '20 at 19:54
  • $\begingroup$ Set $x^2=y$ then use beta function then euler reflection identity. $\endgroup$ Sep 2 '20 at 20:48
  • $\begingroup$ An approach using residues is here. $\endgroup$
    – metamorphy
    Jan 21 '21 at 11:29
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Evaluating $$\oint_C \frac{z^{\alpha-1}}{1+z} dz$$ we see that there is a branch cut along the positive $x-$axis and a pole at $z=-1$.

Take $C$ to be a keyhole contour consisting of a segment from $\epsilon$ to $R$, a circle of radius $R$, a segment from $R$ to $\epsilon$ and a small circle of radius $\epsilon$ surrounding the origin.

The result is:

$$\int_0^\infty \frac{ x^{\alpha-1}}{1+x} = \frac{\pi}{\sin \pi \alpha} \quad \text{when } 0<\alpha<1.$$

With the substitution $y^{1/2} = x$, our integral becomes

$$\bbox[5px, border: 1pt solid blue]{\int_0^\infty \frac{x^{1/3}}{1+x^2} dx = \frac{1}{2} \int_0^\infty \frac{y^{-1/3}}{1+y} dy = \frac{\pi}{2\sin \frac{2\pi}{3}}=\frac{\pi}{\sqrt{3}}.}$$

UPDATE:

In response to J.G.'s question:

The residue at $z=-1$ is $$b=\text{Res}_{z=-1} \frac{z^{\alpha-1}}{1+z}=e^{\pi i (\alpha-1)}.$$

So $$\oint_C \frac{z^{\alpha-1}}{1+z} dz = 2\pi i b$$

On the first segment (from $\epsilon$ to $R$), $z^{\alpha-1}=x^{\alpha-1}$, on the return trip, $z^{\alpha-1}=(e^{2\pi i} x)^{\alpha-1}.$

The integrals along the circles go to zero as $\epsilon\to0$, $R\to0$.

$$\int_0^\infty \frac{x^{\alpha-1}}{1+x}dx - \int_0^\infty \frac{e^{2\pi i (\alpha-1)} x^{\alpha-1}}{1+x}dx = 2\pi i e^{\pi i (\alpha-1)}$$

$$\begin{aligned} \int_0^\infty \frac{x^{\alpha-1}}{1+x}dx &=\frac{2\pi i e^{\pi i (\alpha-1)}}{1-e^{2\pi i (\alpha-1)}}\\ &= \frac{2\pi i}{e^{-\pi i (\alpha-1)}-e^{\pi i (\alpha-1)}} \\ &=\frac{\pi}{\sin \pi(1-\alpha)} \\ &= \frac{\pi}{\sin \pi \alpha}. \end{aligned}$$

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    $\begingroup$ Yes, "keyhole contour" is the critical point. This is one of those iconic problems that is also useful now and then in real life. $\endgroup$ Sep 2 '20 at 19:10
  • $\begingroup$ (+1) The classical keyhole contour works like a charm. $\endgroup$
    – Mark Viola
    Sep 2 '20 at 19:26
  • $\begingroup$ Could you show how this gets us to $\pi\csc(\pi\alpha)$? $\endgroup$
    – J.G.
    Sep 2 '20 at 20:02
  • $\begingroup$ @J.G., yes, edited my answer to show these details. Thanks. $\endgroup$
    – mjw
    Sep 2 '20 at 22:17
  • $\begingroup$ @mjw can you please add more details to your update to J.G. 's question. I am having difficulties in understanding the proof but I really want to understand it as its's really nice. $\endgroup$
    – Avenger
    Sep 19 '20 at 16:06
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With $x=\tan t$ it becomes $\int_0^{\pi/2}\tan^{1/3}tdt$. This can be evaluated in terms of the Beta function. In particualar, $\int_0^{\pi/2}\tan^{2s-1}tdt=\tfrac12\pi\csc\pi s$ implies your integral is $\tfrac12\pi\csc\frac{2\pi}{3}=\frac{\pi}{\sqrt{3}}$.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With Ramanujan's Master Theorem: \begin{align} \int_{0}^{\infty}{x^{1/3} \over 1 + x^{2}}\,\dd x & \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{\infty}{x^{\color{red}{2/3} - 1} \over 1 + x}\,\dd x \end{align} Note that $\ds{{1 \over 1 + x} = \sum_{k = 0}^{\infty}\pars{-1}^{k}x^{k} = \sum_{k = 0}^{\infty}\color{blue} {\Gamma\pars{k + 1}}{\pars{-x}^{k} \over k!}}$.

Then, \begin{align} \int_{0}^{\infty}{x^{1/3} \over 1 + x^{2}}\,\dd x & = {1 \over 2}\,\Gamma\pars{\color{red}{2 \over 3}} \color{blue} {\Gamma\pars{1 + \bracks{-\,\color{red}{2 \over 3}}}} \\[2mm] &= {1 \over 2}\,{\pi \over \sin\pars{\pi/3}} = \bbx{{\root{3} \over 3}\,\pi} \\ & \end{align}

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First, let's do a substitution to reduce the number of poles to be computed. In this problem, it wouldn't be necessary, since the function is fairly easy and there aren't too many poles, however it's a good a practice since in the future you may encounter more intricated problems.

$$\int_{0}^{\infty}{\frac{x^\frac{1}{3}}{x^2+1}dx}=\frac{1}{2}\int_{0}^{\infty}{\frac{z^{-\frac{1}{3}}}{z+1}dz}$$

Now, let’s define our function and integration path. Bear in mind that since we are working with complex functions $z^{-\frac{1}{3}}=\exp{\left(-\frac{Log\left(z\right)}{3}\right)}$, wich means that we must take care of the branch point. Hence, the selection of the keyhole contour. $$f(z)=\frac{z^{-\frac{1}{3}}}{z+1}$$

enter image description here

$$\oint f(z)dz=\left(\int_{0+ir}^{R+ir}+\int_{\Gamma}+\int_{R-ir}^{0-ir}+\int_\gamma \right)f(z)dz$$

1)Let's start working with the two integrals in the positive axis of the Real Line: $$\displaystyle{\lim_{R\rightarrow \infty\\ r\rightarrow 0}}\left(\int_{0+ir}^{R+ir}\int_{R-ir}^{0-ir} \right)f(z)dz$$

Remember that: $$Log(z)=\log|z|+i\text{Arg}(z)$$

$$\int_0^\infty \frac{\exp\left(-\frac{\log\left(z\right)+0i}{3}\right)}{z+1}dz+\int_\infty^0 \frac{\exp\left(-\frac{\log\left(z\right)+2\pi i}{3}\right)}{z+1}dz=\left(1-e^{-\frac{2\pi i}{3}}\right)\int_0^\infty\frac{z^{-\frac{1}{3}}}{z+1}dz$$

  1. To compute the others integrals let's make use of the the Estimation Lemma: $$\left\lvert\int_\Gamma f(z)dz\right\rvert\leq\displaystyle{\lim_{R\rightarrow \infty}}\left\lvert\int_0^{2\pi}\frac{\left(Re^{it}\right)^{-\frac{1}{3}}}{Re^{it}+1}Rie^{it}dt\right\rvert\leq\displaystyle{\lim_{R\rightarrow \infty}}2\pi R^{-\frac{1}{3}}\rightarrow 0$$

$$\left\lvert\int_\gamma f(z)dz\right\rvert\leq\displaystyle{\lim_{r\rightarrow 0}}\left\lvert\int_{2\pi}^{0}\frac{\left(re^{it}\right)^{-\frac{1}{3}}}{re^{it}+1}rie^{it}dt\right\rvert\leq\displaystyle{\lim_{r\rightarrow 0}}2\pi r^{\frac{2}{3}}\rightarrow 0$$

  1. Finally, let's compute the residue: $$\oint f(z)dz=2\pi i\displaystyle{\lim_{z\rightarrow -1}}\frac{z^{-\frac{1}{3}}}{z+1}(z+1)=2\pi ie^{-\frac{\pi i}{3}}$$

Gathering all results: $$\int_0^\infty\frac{z^{-\frac{1}{3}}}{z+1}dz=\frac{2\pi ie^{-\frac{\pi i}{3}}}{1-e^{\frac{-2\pi i}{3}}}=\frac{\pi }{\frac{e^{\frac{\pi i}{3}}-e^{-\frac{\pi i}{3}}}{2i}}=\frac{\pi}{\sin\left(\frac{\pi}{3}\right)}=\frac{2\pi}{\sqrt{3}}$$

Hence: $$\boxed{\int_{0}^{\infty}{\frac{x^\frac{1}{3}}{x^2+1}dx}=\frac{\pi}{\sqrt{3}}}$$

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  • $\begingroup$ Why in point 2 of your answer : when you are making use of estimation lemma, Then how did you estimated 1st inequality in 1st line of 2. I think the inequality must be opposite $\endgroup$
    – Avenger
    Feb 12 '21 at 6:40
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Substitution $$x=y^{3/2}$$ allows to write $$I=\int\limits_0^\infty\dfrac{\sqrt[\large3]x\,\text dx}{1+x^2} = \dfrac32\int\limits_0^\infty\dfrac{y\,\text dy}{1+y^3}.$$ At the same time, substitution $$y=\dfrac1z$$ gives $$I= \dfrac32\int\limits_0^\infty\dfrac{\text dz}{1+z^3} = \dfrac32\int\limits_0^\infty\dfrac{\text dy}{1+y^3}.$$ Therefore, $$I= \dfrac34\int\limits_0^\infty\dfrac{(1+y)\,\text dy}{1+y^3} = \dfrac34\int\limits_0^\infty\dfrac{\text dy}{1-y+y^2} =3\int\limits_0^\infty\dfrac{\text dy}{3+(2y-1)^2}.\tag1$$ Integral $(1)$ does not require the residue approach: $$I=\dfrac{\sqrt3}2\arctan\dfrac{2y-1}{\sqrt3}\bigg|_0^\infty = \dfrac{\sqrt3}2\left(\dfrac\pi2+\dfrac\pi6\right),$$ $$\color{brown}{\mathbf{I=\dfrac\pi{\sqrt3}.}}$$

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