3
$\begingroup$

Compute $\displaystyle\int_0^\infty \frac{dx}{1+x^3}$ by integrating $\dfrac{1}{1+z^3}$ over the contour $\gamma$ (defined below) and letting $R\rightarrow \infty$.

The contour is $\gamma=\gamma_1+\gamma_2+\gamma_3$ where $\gamma_1(t)=t$ for $0\leq t \leq R$, $\gamma_2(t)=Re^{i\frac{2\pi}{3}t}$ for $0\leq t \leq 1$, and $\gamma_3(t)=(1-t)Re^{i\frac{2\pi}{3}}$ for $0\leq t \leq 1$.

So, the contour is a wedge, and by letting $R\rightarrow \infty$ we're integrating over one third of the complex plane. I believe this means we are integrating over the entire complex plane under the substitution $u=x^3$. There are poles at $-\zeta$ for each third root of unity $\zeta$, so there's only one pole in this wedge. I'll just refer to that pole as $-\zeta$.

I guess this means that we can use the residue theorem to say $$\int_{\gamma}\frac{1}{1+z^3}dz=2\pi i\eta(\gamma,-\zeta)\operatorname{Res}\left(\frac{1}{1+z^3},-\zeta\right)=2\pi i \lim_{z\rightarrow -\zeta}\left[(z+\zeta)\frac{1}{1+z^3}\right]$$

I can't evaluate this limit. Also I don't see how it involves $R$, which I'm supposed to be taking a limit of. I suspect I've done something wrong.

What's the problem? How do I proceed?

Also, after I do properly evaluate this integral, I am assuming that its value is supposed to be $\displaystyle\int_0^\infty\frac{dx}{1+x^3}$. Why? (I think I know why conceptually but I need to see how one rigorously writes that out.)

(Note: This is exam review, not homework.)

$\endgroup$

marked as duplicate by YuiTo Cheng, Cesareo, mrtaurho, José Carlos Santos complex-analysis Jun 30 at 19:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $1-5 \to 1-t$ in the contour?! $\endgroup$ – Sharkos Apr 20 '13 at 23:42
  • $\begingroup$ @Sharkos Whoops, yes. $\endgroup$ – Samuel Handwich Apr 20 '13 at 23:44
  • $\begingroup$ Hint : if $g:z\mapsto 1+z^3$, then $(z+\zeta)\dfrac 1{1+z^3}=\dfrac{z-(-\zeta)}{g(z)-g(-\zeta)}$ for all $z$ such that $g(z)\neq 0$. $\endgroup$ – Philippe Malot Apr 20 '13 at 23:48
  • $\begingroup$ @girianshiido Aren't the $z$ for which $g(z)=0$ precisely the ones we're interested in? $\endgroup$ – Samuel Handwich Apr 20 '13 at 23:54
  • $\begingroup$ You said that there are only three such values :) When $z$ is close to $-\zeta$ yet different from it, you know that $g(z)\neq 0$, and that suffices. So what is the limit ? $\endgroup$ – Philippe Malot Apr 20 '13 at 23:56
4
$\begingroup$

Hints: Firstly, $1+z^3 = (z+\zeta)(z+\zeta^2)(z+1)$ by factorizing the polynomial.

Secondly, $$\int _0^1\frac{1}{1+z^3} z'(t) \mathrm d t $$ on the contour $z = R(1-t)\exp(2\pi i/3)$ can be related to the real integral you were originally looking for just by substituting this expression for $z$ in.

$\endgroup$
  • $\begingroup$ Oh god, of course. It is so obvious. Thank you. $\endgroup$ – Samuel Handwich Apr 20 '13 at 23:56
  • $\begingroup$ No worries! Sometimes we convince ourselves it's something we don't understand when it's perfectly simple (: $\endgroup$ – Sharkos Apr 21 '13 at 0:00
1
$\begingroup$

The easier way out to compute $\displaystyle \int_0^{\infty} \dfrac{dx}{1+x^3}$ is as follows. We have $$I = \int_0^{\infty} \dfrac{dx}{1+x^3} = \int_{\infty}^0 -\dfrac1{x^2}\dfrac{dx}{1+1/x^3} = \int_0^{\infty} \dfrac{xdx}{1+x^3}$$ We hence have $$2I = \int_0^{\infty} \dfrac{1+x}{1+x^3} dx = \int_0^{\infty}\dfrac{dx}{1-x+x^2} = \int_0^{\infty} \dfrac{dx}{\left(\dfrac{\sqrt3}2 \right)^2+ \left(x - \dfrac12\right)^2}$$ Hence, we get that $$2I = \left. \dfrac2{\sqrt3}\arctan\left(\dfrac{2x-1}{\sqrt3}\right) \right \vert_0^{\infty} = \dfrac2{\sqrt3}\left(\dfrac{\pi}2 + \dfrac{\pi}6\right) \implies I = \dfrac{2 \pi}{3\sqrt3}$$

$\endgroup$
  • $\begingroup$ I hate to be picky, but that wasn't the question. I am supposed to be using the contour given. This is for a complex analysis class. Thank you anyway, though, sincerely. (I didn't downvote.) $\endgroup$ – Samuel Handwich Apr 20 '13 at 23:46
  • $\begingroup$ @SamuelHandwich No problem. $\endgroup$ – user17762 Apr 20 '13 at 23:49
  • $\begingroup$ That is a very nice development. +1 and you're already too veteran for being worried about downvotes from pety people. $\endgroup$ – DonAntonio Apr 20 '13 at 23:58
  • 1
    $\begingroup$ @DonAntonio :) I am not against down-votes. But down-voting with a reason/ without a comment is what puts me off sometimes. $\endgroup$ – user17762 Apr 21 '13 at 0:01
  • 1
    $\begingroup$ I hate the idea of downvotes without explanation, but if there are two answers, both on 0 votes, say, and one is wrong, I think it's helpful to the OP to distinguish them in this way. I'll remove it if/when it's fixed ofc. $\endgroup$ – Sharkos Apr 21 '13 at 15:06
1
$\begingroup$

$$\frac{1}{x^3+1}=\frac{1}{3(x+1)}-\frac{x-2}{3(x^2-x+1)}$$

But

$$\frac{x-2}{x^2-x+1}=\frac{1}{2}\frac{2x-1}{x^2-x+1}-\frac{\frac{3}{2}}{\frac{3}{4}+\left(x-\frac{1}{2}\right)^2}=\frac{(x^2-x+1)'}{x^2-x+1}-\frac{4}{3}\frac{\frac{3}{2}}{1+\left(\frac{2}{\sqrt3}\left(x-\frac{1}{2}\right)\right)^2}=$$

$$=\frac{(x^2-x+1)'}{x^2-x+1}-\sqrt3\,\frac{\frac{2}{\sqrt3}dx}{1+\left(\frac{2}{\sqrt3}\left(x-\frac{1}{2}\right)\right)^2}$$

Finally:

$$\int\limits_0^\infty\frac{dx}{x^3+1}=\left.\left[\frac{1}{3}\log\frac{\sqrt{x^2-x+1}}{x+1}+\sqrt 3\arctan\frac{2}{\sqrt 3}\left(x-\frac{1}{2}\right)\right]\right|_0^\infty=$$

$$0+\sqrt3\,\left(\frac{\pi}{2}-\arctan\left(-\frac{1}{\sqrt3}\right)\right)=\sqrt3\left(\frac{\pi}{2}+\frac{\pi}{6}\right)=\frac{2\pi}{\sqrt3}$$

$\endgroup$

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