I'm doing this example of Cauchy principle value

$$ \int_0^\infty \frac{dx}{x^3+1}=\frac{2\pi}{3\sqrt{3}} $$

After some steps i got,

$$ \int_{[0,R]+C_R} \frac{dz}{z^3+1}=2\pi i(B_1)\text{ where }B_1= \operatorname{Res}_{z=z_0}\frac{1}{z^3+1} $$

also I got that $\displaystyle \bigg|\int_{C_R} \frac{dz}{z^3+1}\bigg|\to 0 \text{ as } R \to \infty$

There is problem to to finding residue at $z_0=\displaystyle \frac{1+\sqrt{3}i}{2}$

Here i am considering the following contour:

enter image description here

please help me.thanks in advance.


The contour is good. Two things though:

1) You have to consider the integral along the angled line of the wedge contour. The angle of the contour was chosen to preserve the integrand. 2) Write $z=e^{i 2 \pi/3} x$ and get that the contour integral is

$$\left(1-e^{i 2 \pi/3}\right) \int_0^{\infty} \frac{dx}{x^3+1} = i 2 \pi \frac{1}{3 e^{i 2 \pi/3}}$$

The term on the right is the residue at the pole $z=e^{i\pi/3}$ times $i 2\pi$. I used the fact that, if $f(z)=a(z)/b(z)$, then the residue of a simple pole $z_k$ of $f$ is $a(z_k)/b'(z_k)$.

Note that $e^{i 2 \pi/3}-e^{i 4 \pi/3}=i \sqrt{3}$. The result follows.


$$B_1= \operatorname{Res}\limits_{z=z_0}\frac{1}{z^3+1}=\lim\limits_{z\to {z_0}}{\left(\frac{1}{z^3+1}(z-z_0)\right)}=\lim\limits_{z\to {z_0}}{\left(\frac{z-z_0}{(z+1)(z-z_0)(z-\bar{z}_0)}\right)}=\frac{1}{(z_0+1)(z_0-\bar{z}_0)}=\frac{1}{\left(\dfrac{1+\sqrt{3}i}{2} +1 \right)}.$$ Contour of integration $\gamma$ consists of three parts: $\gamma=\gamma_1^{+}\cup C_R^{+} \cup \gamma_2^{+}$ $$ \gamma_1=[0,\,R]=\{z=\rho e^{\frac{2\pi i}{3}}, \;\;{0}\leqslant {\rho}\leqslant{R} \}, \\ \gamma_2=\{z=\rho e^{\frac{2\pi i}{3}}, \;\;{0}\leqslant {\rho}\leqslant{R} \}.$$ Thus $$ \int\limits_{\gamma_1^{+}\cup C_R^{+} \cup \gamma_2^{+}} \frac{dz}{z^3+1}=2\pi i B_1. $$ You missed integral $$ \int\limits_{\gamma_2^{+}} \frac{dz}{z^3+1}=-\int\limits_{R}^{0} \frac{ e^{\frac{2\pi i}{3}}d{\rho}}{{\rho}^3+1}=-e^{\frac{2\pi i}{3}} \int\limits_{0}^{R} \frac{d{\rho}}{{\rho}^3+1}. $$

  • $\begingroup$ I can't get exact ans. which i want. i can't cancel $i$ from denominator of residue and $2\pi$i. $\endgroup$ – Siddhant Trivedi Mar 23 '13 at 6:03

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