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I know this has been answered on the site (see here: How to integrate $\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$?) but I want to know why what I did is not leading to a sensible result.

I used the splitting the logarithm trick, i.e. examining $$ \int_\Gamma\frac{z^{1/3}\log z}{1+z^2}\mathrm dz=2\pi i\left( \frac{\sqrt{3}}{2}+\frac{i}{2}\right )=\pi(\sqrt{3} i-1) $$ by the residue theorem and using the sane branch cut on the positive real axis. Now by Jordan's theorem the contribution of the arc should be zero leaving us with $$ \int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx+\int_{-R}^0\frac{x^{1/3}\log x}{1+x^2}\mathrm dx\\ = \int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx+\int_{0}^R\frac{x^{1/3}\log(- x)}{1+x^2}\mathrm dx\\ =2\int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx+\frac{\pi}{2}i\int_0^R\frac{x^{1/3}}{1+x^2}\mathrm dx $$ which by equating real and imaginary parts should imply my integral is $$ 2\int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx+\frac{\pi}{2}i\int_0^R\frac{x^{1/3}}{1+x^2}\mathrm dx=\pi(\sqrt{3} i-1)\\ \implies \int_0^\infty\frac{x^{1/3}}{1+x^2}\mathrm dx=2\sqrt{3} $$ which is false. Additionally, this would imply that $$ \int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx=-\frac{\pi}{2} $$ which is also false.

Where did I go wrong? My guess is somewhere in the branch cutting, possibly with taking the cube root of 3 in the residue theorem?

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  • $\begingroup$ You do not need the logarithm. The power term provides the branch point that you need to derive the integral in terms of the residues of the poles at $\pm i$. $\endgroup$ – Ron Gordon Jul 17 '17 at 6:19
  • $\begingroup$ @RonGordon as in the linked answer? $\endgroup$ – qbert Jul 17 '17 at 6:19
  • $\begingroup$ Yes, but the solution can be simplified. $\endgroup$ – Ron Gordon Jul 17 '17 at 6:21
  • $\begingroup$ thank god... it's currently unpleasant to look at. Do you mind explaining what you mean by "the power term provides the branch point" $\endgroup$ – qbert Jul 17 '17 at 6:22
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To do: Integration along the circle sector $\,\Gamma_R\,$ through the origin

and limitation by the corner points $\,R>0\,$ and $\,e^{i\pi}R=-R$ .

The $log$ in your integral is unnecessary and makes the calculation error-prone.

Hint: Wolframalpha tells that $\,\displaystyle\int\limits_0^\infty \frac{x^{1/3}\ln x}{1+x^2}=\frac{\pi^2}{6}\,$ and $\,\displaystyle\int\limits_{-\infty}^0 \frac{x^{1/3}\ln x}{1+x^2}=-\frac{\pi^2}{12}(5-i3\sqrt{3})$ .

With other words: $\,\displaystyle\int\limits_\Gamma \frac{z^{1/3}\ln z}{1+z^2}dz\,$ contents $\,\pi^2\,$, not only $\,\pi\,$. (I haven't calculated this integral but obviously you have a wrong result for this).


Only to see, that it's easier to calculate without any tricks.

We get on the one hand

$\displaystyle \lim\limits_{R\to\infty} \oint_{\Gamma_R}f dz=i2\pi \cdot\text{res}\left( \frac{z^{1/3}}{1+z^2},z=i \right)= -i\pi e^{i2\pi/3} $

and on the other hand

$\displaystyle \lim\limits_{R\to\infty} \oint_{\Gamma_R}f dz=\int_{\mathbb{R}^+}f dz -\int_{e^{i\pi} \mathbb{R}^+}f dz=\int_{\mathbb{R}^+}f(z) dz -\int_{\mathbb{R}^+}f(e^{i\pi} z) e^{i\pi} dz$

$\displaystyle =\int_{\mathbb{R}^+}f(z) dz - e^{i2\pi(2/3)} \int_{\mathbb{R}^+}f(z)dz= (1- e^{i4\pi/3}) \int_{\mathbb{R}^+}f(z) dz $ .

It follows:

$\displaystyle \int_{\mathbb{R}^+}f(z) dz =\frac{-i\pi e^{i2\pi/3} }{1-e^{i4\pi/3}}=\frac{\pi}{\sqrt{3}}$

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