Evaluation of definite integral using residue theorem：$ \int^{+\infty}_{-\infty} \frac{x-1}{x^3-1} dx$

Question

$$ \int^{+\infty}_{-\infty} \frac{x-1}{x^3-1} dx$$

I need to evaluate the above integral .

My idea is to consider the same integral but with the $x$'s as $z$'s, over the complex plane, have a closed contour integral over $\gamma$, and then use the residue theorem. i.e. consider:

$$ \int^{+\infty}_{-\infty} \frac{z-1}{z^3-1} dz$$

I'm stuck on how to formulate $\gamma$ though.

I know this has 3 poles: at

$z=1$, $z= \frac{-1}{2} + i\frac{\sqrt3}{2}$ and $z= \frac{-1}{2} - i\frac{\sqrt3}{2}$

How do I use this to divide up gamma over contours to which I can then use the residue theorem? And then do I have to either evaluate directly or apply the ML inequality to each individual contour?

Alex Youcis · Answer 1 · 2013-03-09T22:34:25.603

1

This seems quite silly to use complex analysis for.

$$\begin{aligned}\int_{-\infty}^{\infty}\frac{x-1}{x^3-1}\,dx &=\int_{-\infty}^\infty \frac{dx}{1+x+x^2}\\ &= \frac{2}{\sqrt{3}}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\mid_{-\infty}^{\infty}\\ &= \frac{2\pi}{\sqrt{3}}\end{aligned}$$

If you're hell-bent on using the Residue Theorem though, let $\Gamma_R$ be the semi-cicular contour in the upper half-plane which is centered at $0$ and of radius $R$. Then, you know that

$$\int_{\Gamma_R}\frac{1}{z^2+z+1}\,dz = \int_{-R}^{R}\frac{dx}{1+x+x^2}+\int_{A_R}\frac{dz}{1+z+z^2}$$

where $A_R$ is the upper circular arc of $\Gamma_R$. Now, $\displaystyle \frac{1}{1+z+z^2}$ has only one pole in the interior of $\Gamma_R$, at $\displaystyle z_0=\frac{-1}{2}+i\frac{\sqrt{3}}{2}$.

Thus, you see that for every $R$ the Residue Theorem gives

$$\int_{\Gamma_R}\frac{dz}{1+z+z^2}=2\pi i\, \text{Res}\left(\frac{1}{1+z+z^2},z_0\right)$$

Thus, if you can show that

$$\displaystyle \int_{A_R}\frac{dz}{z^2+z+1}\xrightarrow{R\to\infty}0$\quad\mathbf{(1)}$$

You will have that

$$\int_{-\infty}^{\infty}\frac{dx}{1+x+x^2}=2\pi i\, \text{Res}\left(\frac{1}{1+z+z^2},z_0\right)$$

The residue, as I've said, is up to you. To prove $\mathbf{(1)}$ let me give you a hint: Jordan's Lemma.

edited Mar 09 '13 at 22:34

answered Mar 09 '13 at 22:24

Alex Youcis

54,059

What if the $x^3$ term was to a higher power though? – Sapph Mar 09 '13 at 22:28
@Sapph See my edit. – Alex Youcis Mar 09 '13 at 22:38
Jordan's lemma seems like overkill. – mrf Mar 09 '13 at 22:43
@mrf In this specific case. But Sapph's question seems to make me think they are asking this as a particular example--they want to be able to prove such problems in general. But yes, Jordan's lemma is overkill in this specific case. – Alex Youcis Mar 09 '13 at 22:43
You're right. The actual question is for $x^n$, but I want to solve it for some values of n such that it isn't a trivial integral. I don't understand how to create the contours though – Sapph Mar 09 '13 at 22:51
@Sapph So, you're trying to do $\frac{x-1}{x^n-1}$? – Alex Youcis Mar 09 '13 at 22:52
Yes indeed I am! – Sapph Mar 09 '13 at 23:04
@Sapph You can use the same contour. Your poles are now just at the $n^{\text{th}}$-roots of unity which lie in the upper half-plane. – Alex Youcis Mar 09 '13 at 23:05
Ok, thanks! Would that include the roots that lie on the real axis? – Sapph Mar 09 '13 at 23:12
@Sapph What I said only works for even $n$, since we can't have roots on the real axis. – Alex Youcis Mar 09 '13 at 23:14

score 0 · Answer 2 · answered Mar 09 '13 at 22:45

Note that $\frac{x - 1}{x^3 - 1} = \frac{1}{x^2 + x + 1}$. So the integral is $$ \int_{-\infty}^{+\infty} \frac{1}{x^2 + x + 1} \, dx. $$ First take the integral from $-a$ to $a$ and then take the limit after. Just complete the square in the denominator and use substitution to get an integral in terms of $\arctan$. $$ \int_{-a}^{a} \frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}} \, dx \\ =\frac{4}{3}\int_{-a}^{a} \frac{1}{\left(\frac{2}{\sqrt{3}}(x + \frac{1}{2})\right)^2 + 1} \, dx. $$ Using the substitution $u = \frac{2}{\sqrt{3}}(x + \frac{1}{2})$, we have $dx = \frac{\sqrt{3}}{2}du$ and so the indefinite integral is $$ \frac{4}{3}\frac{\sqrt{3}}{2}\int \frac{1}{u^2 + 1} \, dx \\ =\frac{2}{\sqrt{3}}\arctan(u) = \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2x + 1)\right). $$ Thus, the definite integral is $$ \left.\frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2x + 1)\right)\right|_{-a}^a = \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2a + 1)\right) - \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(-2a + 1)\right). $$ As $a \to \infty$, we have $\arctan\left(\frac{1}{\sqrt{3}}(2a + 1)\right) \to \frac{\pi}{2}$ and $\arctan\left(\frac{1}{\sqrt{3}}(-2a + 1)\right) \to -\frac{\pi}{2}$. Hence the integral is $$ \int_{-\infty}^{+\infty} \frac{x - 1}{x^3 - 1} \, dx= \frac{2}{\sqrt{3}} \cdot \frac{\pi}{2} + \frac{2}{\sqrt{3}} \cdot \frac{\pi}{2} = \frac{2\pi}{\sqrt{3}}. $$

score 0 · Answer 3 · answered Mar 09 '13 at 22:48

If you really insist in complex analysis, define:

$$f(z):=\frac{z-1}{z^3-1}=\frac{1}{z^2+z+1}\,\,,\,\,C_R:=[-R,R]\cup\Gamma_R:=\{z\in\Bbb C\;;\;|z|=R\,,\,\,\Im(z)\ge 0\}$$

so that we have the integral

$$\oint\limits_{C_R}f(z)\,dz=\int\limits_{-R}^R\frac{dx}{x^2+x+1}+\int\limits_{\Gamma_R}f(z)\,dz$$

Now, we have $\,\displaystyle{z^2+z+1=0\iff z=\frac{-1\pm\sqrt 3\,i}{2}=e^{\pm\frac{2\pi i}{3}}}\,$ , but since we're working only on the upper half plane, we only have one simple pole there:

$$Res_{z=e^{2\pi i/3}}(f)=\lim_{z\to e^{2\pi i/3}}\,\left(z-e^{\frac{2\pi i}{3}}\right)f(z)\stackrel{\text{for ex., L'Hospital}}=$$

$$=\frac{1}{2e^{\frac{2\pi i}{3}}+1}=\frac{1}{\sqrt 3\,i}$$

Also, by Cauchy's Evaluation Theorem:

$$\left|\,\,\int\limits_{\Gamma_R}f(z)\,dz\,\,\right|\le\frac{1}{R^2-R-1}\cdot\pi R\xrightarrow [R\to\infty]{}0$$

So by Cauchy's residue Theorem and passing to the limit:

$$\frac{2\pi}{\sqrt 3}=2\pi i\cdot\frac{1}{\sqrt 3\,i}=\lim_{R\to\infty}\int\limits_{C_R}f(z)\,dz=\int\limits_{-\infty}^\infty\frac{dx}{x^2+x+1}$$

Evaluation of definite integral using residue theorem：$ \int^{+\infty}_{-\infty} \frac{x-1}{x^3-1} dx$

3 Answers3