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I'm trying to evaluate this integral using contour integration (over a Riemann surface), but I'm stuck at the step where I need to calculate the residues. The roots of $1+z^{3/2}$ are $1$ and $e^{2\pi*i/3}$, but I don't see any way to extract poles or whatnot from this information, as $z^{1/2}$ seems inseparable from $1+z^{3/2}$.

Help is appreciated!

kiss my armpit
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  • How about using $1+x^{3/2}=(1+x^{1/2})(x-x^{1/2}+1)$? – Ben Grossmann Aug 28 '13 at 17:12
  • How would that help? – row Aug 28 '13 at 17:51
  • I was thinking that the second term gives you something integrable along the real contour, but it doesn't converge at infinity so maybe that isn't so helpful after all. I'll be honest, I'm not really sure what "separating $z^{1/2}$ from $1+z^{3/2}$" refers to here. – Ben Grossmann Aug 28 '13 at 17:55
  • Usually for integrating things like this, you'd bring the expression to something like $x^a/P(x)$, where $P(z)$ is some polynomial in whole powers. That way it's possible to plug $z^{a}=e^{alog(z)}$ into the contour integral (usually some sort of circle minus the real line on the Riemann surface of $log(z)$), and then we can evaluate the expression, making a distinction between the various branches of $log(z)$. – row Aug 28 '13 at 18:11
  • I'm open for seeing other methods, though... – row Aug 28 '13 at 18:12
  • Why not just use $u$-substitution then? Set $u=z^{1/2}$ to get the new integral $$\int_0^\infty \frac{2u,du}{1+u^3}$$ – Ben Grossmann Aug 28 '13 at 18:20
  • Substitute away √x to get a cubic degree in the denominator. Would that be helpful? – imranfat Aug 28 '13 at 18:21
  • Yes, it would! I don't think there's any problem evaluating that integral.

    Still, I'd rather see this done on a Riemann surface, assuming that is possible.

     – row Aug 28 '13 at 18:25

3 Answers3

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First, sub $x=y^2$ to get that the integral is equal to

$$2 \int_0^{\infty} dy \frac{y}{1+y^3}$$

Now we can proceed straightforwardly. Consider the contour integral

$$\oint_C dz \frac{z \, \log{z}}{1+z^3}$$

where $C$ is a keyhole contour about the positive real axis. In this case, one may show that this contour integral is simply

$$-i 2 \pi \int_0^{\infty} dy \frac{y}{1+y^3}$$

and is also equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand, or $z=e^{i \pi/3}$, $z=e^{i \pi}$, and $z=e^{i 5 \pi/3}$. This sum is equal to

$$\frac{e^{i \pi/3} (i \pi/3)}{3 e^{i 2 \pi/3}} + \frac{e^{i \pi} (i \pi)}{3 e^{i 2 \pi}}+\frac{e^{i 5 \pi/3} (i 5 \pi/3)}{3 e^{i 10 \pi/3}}$$

which, upon simplifying, is seen to be equal to

$$i \frac{\pi}{9} (i 2 \sqrt{3}) = -\frac{2 \pi}{3 \sqrt{3}}$$

Therefore the orginal integral is

$$\int_0^{\infty} \frac{dx}{1+x^{3/2}} = \frac{4 \pi}{3 \sqrt{3}}$$

Ron Gordon
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Here's another approach. Consider the general integral

$$\int_0^{\infty} \frac{dx}{1+x^a}$$

where $a>1$. Sub $y=x^a$ and transform the integral to

$$\frac{1}{a} \int_0^{\infty} dy \frac{y^{-(a-1)/a}}{1+y}$$

This integral may be evaluated by considering the contour integral

$$\oint_C dz \frac{z^{-(a-1)/a}}{1+z}$$

where $C$ is a keyhole contour about the positive real axis. One may then show that this integral is equal to

$$\left ( 1-e^{-i 2 \pi (a-1)/a}\right ) \int_0^{\infty} dy \frac{y^{-(a-1)/a}}{1+y} $$

which in turn is equal to $i 2 \pi$ times the residue at $z=e^{i \pi}$, or

$$i 2 \pi e^{-i \pi (a-1)/a}$$

Therefore the integral is

$$\int_0^{\infty} \frac{dx}{1+x^a} = \frac{\pi/a}{\sin{(\pi/a)}}$$

Plugging in $a=3/2$, I get $(2 \pi/3)/\sin(2\pi/3) = 4 \pi/(3 \sqrt{3})$.

Ron Gordon
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As shown in this answer, $$ \int_0^\infty\frac{x^n}{1+x^m}\mathrm{d}x=\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right) $$ Setting $n=0$ and $m=3/2$, we get $$ \begin{align} \int_0^\infty\frac{1}{1+x^{3/2}}\mathrm{d}x &=\frac{2\pi}{3}\csc\left(\frac{2\pi}{3}\right)\\ &=\frac{4\pi}{3\sqrt3} \end{align} $$

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