3

Use the residue theorem to compute:$$I=\int_0^\infty\frac{1}{x^3+1}\,\mathrm{d}x$$By using the contour that goes first from $0\to r$ along the real axis ($\gamma_1$), then in an arc from $r\to r\exp(\frac{2}{3}\pi i)$ ($\gamma_2$), then in a straight line from $r\exp(\frac{2}{3}\pi i)\to0$ ($\gamma_3$), taking limits as $r\to\infty$.

Firstly we need to show that the integral along this contour is equivalent to the integral along the real axis, as $r\to\infty$, so it first must be shown that:

$$\lim_{r\to\infty}\int_{\gamma_2}\frac{1}{z^3+1}\,\mathrm{d}z=0$$

This part is straightforward: noticing that the arc, when cubed, is transformed to a circle of radius $r^3$ - missing the point $(r^3,0)$ - and adding $1$ to all values on this circle shifts the circle such that the smallest value of $|z^3+1|$ is $r^3-1$:

$$0\le\lim_{r\to\infty}\left|\int_{\gamma_2}\frac{1}{z^3+1}\,dz\right|\le\lim_{r\to\infty}\frac{2}{3}\pi r\cdot\sup_{\gamma_2}\frac{1}{|z^3+1|}\le\lim_{r\to\infty}\frac{2}{3}\pi r\cdot\frac{1}{r^3-1}=0$$

So the contour $\gamma_2$ is accounted for. Sadly, using the same estimation trick on the straight line segment $\gamma_3$ fails, since the maximum value of $(z^3+1)^{-1}$ on $\gamma_3$ is just $1$, which fails to give a limit to zero. Deciding to crack the integral by hand, I proceeded:

$$\begin{align}\int_{\gamma_3}\frac{1}{z^3+1}\,\mathrm{d}z&=\int_r^0\frac{\exp\frac{2}{3}\pi i}{(t\exp\frac{2}{3}\pi i)^3+1}\,\mathrm{d}t\\&=-\exp\frac{2}{3}\pi i\int_0^r\frac{1}{t^3+1}\,\mathrm{d}t\end{align}$$

And got immediately stuck. According to Wolfram Alpha, the antiderivative does exist but it is not at all clear that as $r\to\infty$, the integral goes to $0$. Besides, this is a practice problem; although university level, I doubt one has to crack a very tricky antiderivative and an unclear limit. What am I doing wrong? As far as I understand it, in order for the residue theorem to be of any use, the contour integral enclosing the singularity must first be shown to evaluate to the integral along the real axis.

FShrike
  • 40,125
  • Notice that the integral over $\gamma_3$ is a multiple of your original integral. – Pax Sep 20 '21 at 17:41
  • @pax - thank you, that solves it immediately! – FShrike Sep 20 '21 at 17:43
  • You shouldn't modify your question by adding at the beginning instead of at the end how you have solved your issue : as it is at present, it's hard to a (new) reader to understand what was the question... – Jean Marie Sep 20 '21 at 17:50
  • 1
    @JeanMarie I've placed it at the end now, thank you. I thought the common practice was to do the opposite, but clearly I was mistaken – FShrike Sep 20 '21 at 17:51
  • The key here is that the $\frac{1}{z^3 + 1}$ part is the same, but the $dz$ part is not the same (and is off by a multiple $e^{2\pi i / 3})$. Always be careful with $u$-substitutions when you're trying to relate two integrals. – Mark Saving Sep 20 '21 at 18:11
  • This question is solved here. 2) Thanks for having taken into account my remark.
    • – Jean Marie Sep 20 '21 at 18:40
  • @JeanMarie Thank you for the link to the correct solution ... I spent far longer on this question than I should have done, but at least it's done and I know it's right. – FShrike Sep 20 '21 at 20:38
  • Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$ should handle this question. My answer uses contour integration. – robjohn Sep 20 '21 at 21:02
  • @robjohn Incredible! Thank you for this, I will study it. It is not clear to me how the keyhole contour is parametrised - is it like a Hankel contour? – FShrike Sep 20 '21 at 21:06
  • The circular parts can be parametrized as $re^{it}$ for fixed $r$ and $t$ just inside $[0,2\pi]$, and the linear parts are just lines. It shouldn't matter exactly how, but if you really need it, I can work out the parameterization. – robjohn Sep 20 '21 at 21:09
  • I gather one limits it as tight as possible, with the keyhole radius being made less than $\varepsilon$ - is this right? As far exact details, that's fine - as soon as I finally thrashed this one out I wondered if one could use contour techniques for the general integral, and it's been very helpful to see your answer @robjohn – FShrike Sep 20 '21 at 21:11
  • 1
    Here is how I usually specify contours, using $\epsilon\lt r\lt R$: $$\overbrace{\left\sqrt{r^2-\epsilon^2},\sqrt{R^2-\epsilon^2}\right+(0,\epsilon)}^\text{line over the real axis}\cup\overbrace{Re^{i\left[\sin^{-1}(\epsilon/R),2\pi-\sin^{-1}(\epsilon/R)\right]}}^\text{large circle}\\cup\underbrace{\left\sqrt{R^2-\epsilon^2},\sqrt{r^2-\epsilon^2}\right+(0,-\epsilon)}\text{line under the real axis}\cup\underbrace{re^{i\left[2\pi-\sin^{-1}(\epsilon/r),\sin^{-1}(\epsilon/r)\right]}}\text{small circle}$$ each $[a,b]$ range is a range of real numbers between the two values. – robjohn Sep 20 '21 at 21:21