I am having a hard time solving the following integral using the Residue theorem and would appreciate some help:
$$I=\int_0^\infty \frac{dx\sqrt{x}}{1+x^3}$$
There is a branch cut and so the integration must be carried out around it, thus:
$I_1$ is the desired integral; $I_2$ is the integral along the contour as $R\to\infty$ and hence equals zero; $I_3$ is the integration below the positive x-axis from $R\to\infty$ to zero, and is also equal to the desired integral; $I_4$ is the integral along an infinitesimal contour around the branch point $x=0$ and hence equals zero.
Assuming this is correct, $$2\pi i \sum \operatorname{Res}[e^{5\pi i/3}, e^{\pi i/3}, e^{\pi i}] = 2I$$
However, the sum of the residues should be $\frac{-i}{6}$ and unfortunately I am unable to obtain that. As mentioned, I'd appreciate some advice.
Complex Analysis – Complex Analysis and Residue Theorem
complex-analysis
Best Answer
Indeed, it is correct. It's a standard keyhole contour integration path
$\hspace{4cm}$
with the two contributing parts above and below the branch cut on the positive real axis. The change of direction and the branch of the square root each contribute a factor $-1$ for the part below the real axis, so that part of the contour integral equals the part above the real axis, whence the integral is $\dfrac{1}{2}\cdot 2\pi i$ times the sum of the residues.
Another method - simpler in a way, since it requires only the computation of one residue - is to use the boundary of an angular sector with angle $2\pi/3$ as the contour, similar to
$\hspace{4cm}$
but with a greater angle of $2\pi/3$ so that the denominator attains the same values on the rays, and with a small circular arc around the branch point of $\sqrt{z}$ in $0$. The square root produces a factor of $e^{\pi i/3}$ then, due to $\sqrt{e^{2\pi i/3}x} = e^{\pi i/3}\sqrt{x}$, the direction of the ray produces a factor $e^{2\pi i/3}$, due to $d(e^{2\pi i/3}x) = e^{2\pi i/3}\,dx$, and the orientation produces a factor of $-1$ in relation to the part of the contour integral on the real axis. The product of these factors is $1$, so in the limit, the contour integral is again twice the integral over the positive real axis.
Since the value of the integral is $\dfrac{\pi}{3}$, the sum of the residues should be $-\dfrac{i}{3}$, and not $-\dfrac{i}{6}$. And it is. Choosing the branch of the square root with $0 < \arg \sqrt{z} < \pi$ on $\mathbb{C}\setminus [0,\infty)$, we find
$$\operatorname{Res} \left(\frac{\sqrt{z}}{z^3+1}; e^{i\varphi}\right) = \frac{e^{i\varphi/2}}{3e^{2i\varphi}} = \frac{1}{3}e^{-3i\varphi/2}$$
for $e^{i\varphi} = -1$.
In $e^{i\pi (2k+1)/3}, \; k \in \{0,1,2\}$, we thus have
$$\operatorname{Res} \left(\frac{\sqrt{z}}{z^3+1}; e^{i\pi (2k+1)/3}\right) = \frac{1}{3} e^{- i\pi (2k+1)/2} = \frac{(-1)^k}{3i},$$
and the sum of the residues is $\frac{1}{3i}$, whence
$$\int_0^\infty \frac{\sqrt{x}}{x^3+1}\,dx = \pi i \cdot \frac{1}{3i} = \frac{\pi}{3}.$$
The images for the contours shamelessly taken from this and this answer on the site.