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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.

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  • $\begingroup$ I am quite positive the method delineated above is correct. The problem is obtaining the correct value for the sum of the residues. With that I'd appreciate help. $\endgroup$ – Grtv Jan 20 '14 at 14:23
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I am quite positive the method delineated above is correct.

Indeed, it is correct. It's a standard keyhole contour integration path

$\hspace{4cm}$keyhole contour

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

$\hspace{4cm}$sector contour

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.

The problem is obtaining the correct value for the sum of the residues. With that I'd appreciate help.

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.

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  • $\begingroup$ Thank you so much! Is there any way you could attach a diagram depicting the integration path? It would render the explanation far more lucid, as far as I am concerned, and would be most appreciated. $\endgroup$ – Grtv Jan 20 '14 at 16:10
  • $\begingroup$ I'm terrible with pictures, I'll see if I can figure out how to draw one, but I make no promises. $\endgroup$ – Daniel Fischer Jan 20 '14 at 16:16
  • $\begingroup$ Here's a suggestion - perhaps you could find a diagram online which represents to some extent the same rationale and provide the link. $\endgroup$ – Grtv Jan 20 '14 at 16:32
  • $\begingroup$ For your path, the search term would be "keyhole contour". That's pretty standard, should turn up a few items (I guess there's a good chance that you can search better than I, I'll try nevertheless). For the sector, I'm not sure what the terms should be. Let's see if I can find something. $\endgroup$ – Daniel Fischer Jan 20 '14 at 16:39
  • $\begingroup$ I've found some contours on site, hope the pictures help. If not, tell me what I could explain better. $\endgroup$ – Daniel Fischer Jan 20 '14 at 17:09

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