Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
$1-5 \to 1-t$ in the contour?! –  Sharkos Apr 20 '13 at 23:42
    
@Sharkos Whoops, yes. –  Samuel Handwich Apr 20 '13 at 23:44
    
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 $g(z)\neq 0$. –  Philippe Malot Apr 20 '13 at 23:48
    
@girianshiido Aren't the $z$ for which $g(z)=0$ precisely the ones we're interested in? –  Samuel Handwich Apr 20 '13 at 23:54
    
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 ? –  Philippe Malot Apr 20 '13 at 23:56

3 Answers 3

up vote 3 down vote accepted

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.

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

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 $$2I = \left. \dfrac2{\sqrt3}\arctan\left(\dfrac{2x-1}{\sqrt3}\right) \right \vert_0^{\infty} = \dfrac2{\sqrt3}\left(\dfrac{\pi}2 + \dfrac{\pi}6\right) \implies I = \dfrac{2 \pi}{3\sqrt3}$$

share|improve this answer
    
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.) –  Samuel Handwich Apr 20 '13 at 23:46
    
@SamuelHandwich No problem. –  user17762 Apr 20 '13 at 23:49
    
That is a very nice development. +1 and you're already too veteran for being worried about downvotes from pety people. –  DonAntonio Apr 20 '13 at 23:58
1  
@DonAntonio :) I am not against down-votes. But down-voting with a reason/ without a comment is what puts me off sometimes. –  user17762 Apr 21 '13 at 0:01
1  
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. –  Sharkos Apr 21 '13 at 15:06

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

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:

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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