I am asked to integrate: $$\int_0^\infty \frac{1}{(1+x)x^{1/3}} dx$$

Complexification of this integral leads to: $$f(z)=\frac{1}{(1+z)z^{1/3}}$$ singularities: $z=0$ and $z=-1$. So I thought, let's make it easy and pick $\frac{-\pi}{2}<arg (z)<\frac{3\pi}{2}$ and use this contour (let's name it $C$).


This contour doesn't have any singularities, hence (by Cauchy-Goursat theorem): $$\oint_C f(z)dz=0$$$$=[\oint_A+\oint_F+\oint_B+\oint_D]f(z)dz$$

Because $\oint_F f(z)dz=0$ when 'R' goes to infinity (i.e. F 'blows' up) and $\oint_D f(z)dz$ goes to zero when '$\epsilon$' goes to zero, we are left with: $$[\oint_A+\oint_B]f(z)dz=0$$ Parametrisation results for $A$ in: $z=x$ from $\epsilon$ to $R$, parametrisation for $B$ results in $z=xe^{i\pi}$ from $R$ to $\epsilon$.

We obtain: $$\oint_A f(z)dz=\int_\epsilon^R \frac{1}{(1+x)x^{1/3}} dx$$ and, because $z^{1/3}=exp(\frac{1}{3}(log(z)\cdot i arg(z))$ $$\oint_B f(z)dz=-e^{-i\frac{\pi}{3}}\int_\epsilon^R \frac{1}{(1-x)x^{1/3}} dx$$ with $R\to\infty$ and $\epsilon\to 0$.

We can't conclude anything about this integral. Could anyone explain me why this contour doesn't work?

  • 4
    $\begingroup$ In your illustrate the arc $B$ of the contour $C$ appears to pass through the pole at $z = -1$. $\endgroup$ Nov 2 '15 at 17:13
  • 1
    $\begingroup$ if you avoid $-1$ by a small semicircle (like you have done with zero) you may calculate your integral $\endgroup$
    – tired
    Nov 2 '15 at 17:48

You just provided a terrific service to all of those who wish to learn how to use contour integration to evaluate definite integrals. Really, you answered your own question. The contour you chose is not useful in evaluating the sought-after integral because you wound up with another integral equally difficult to evaluate.

The goal of using contour integration is to express the desired definite integral in terms of other quantities that are known, or at least easier to evaluate. In your case, you wanted to evaluate

$$\int_0^{\infty} dx \, \frac{x^{-1/3}}{x+1} $$

but you wound up expressing that integral in terms of another:

$$e^{-i \pi/3} PV \int_0^{\infty} dx \, \frac{x^{-1/3}}{x-1} + \pi \, e^{i \pi/6} $$

(NB I corrected your incorrect result - you need a detour about a pole on the contour.)

This is of little help and is a red flag that you need a different contour. One way to go is to substitute as @Jack has done. However, if it is important to evaluate directly using contour integration, then a smarter choice of contour is a keyhole contour about the positive axis. In this case, $\arg{z} \in [0,2 \pi]$. The contour integral is then (in the limits as the outer radius goes to $\infty$ and the inner radius goes to 0)

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

By the residue theorem, the contour integral is $i 2 \pi$ times the residue of the pole of the integrand at $z=e^{i \pi}$, so that

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


The main issue is that $z^{1/3}$ is not a holomorphic function or meromorphic function, it has a branch point in the origin. Anyway, by setting $x=z^3$ we have: $$ \int_{0}^{+\infty}\frac{dx}{(1+x)x^{1/3}} = 3\int_{0}^{+\infty}\frac{z\,dz}{1+z^3} $$ and without using complex analysis, but just splitting the integration range in two intervals and applying the change of variable $z\mapsto\frac{1}{z}$ in the "rightmost" integral: $$ \int_{0}^{+\infty}\frac{z}{1+z^3}\,dz = \int_{0}^{1}\frac{z\,dz}{1+z^3}+\int_{0}^{1}\frac{dz}{z^3+1} = \int_{0}^{1}\frac{dz}{z^2-z+1} $$ we have: $$ \int_{0}^{+\infty}\frac{dx}{(1+x)x^{1/3}} = 2\sqrt{3}\,\left.\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\right|_{0}^{1}=\color{red}{\frac{2\pi}{\sqrt{3}}}.$$


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