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?

  • $\begingroup$ 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

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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.