How to evaluate this by contour integral?

Question

I want to evaluate cauchy principal value by complex contour integral, but I failed. Is this integral solved by contour integral? $$PV\int_0^\infty\frac{1}{x^3-1}\,\mathrm{d}x$$

Answer 1

If you are not forced to use contour integral, then we may write

$$ \frac{1}{x^3 - 1} = \frac{1}{3(x-1)} - \frac{x+2}{3 (x^2+x+1)} $$

and decompose OP's principal value as

\begin{align*} \text{PV}\!\! \int_{0}^{\infty} \frac{1}{x^3 - 1} \, \mathrm{d}x &= \lim_{R\to\infty} \text{PV}\!\! \int_{0}^{R} \frac{1}{x^3 - 1} \, \mathrm{d}x \\ &= \lim_{R\to\infty} \biggl[ \text{PV}\!\! \int_{0}^{2} \frac{1}{x^3 - 1} \, \mathrm{d}x + \int_{2}^{R} \frac{1}{x^3 - 1} \, \mathrm{d}x \biggr] \\ &= \lim_{R\to\infty} \biggl[ \int_{2}^{R} \frac{1}{3(x-1)} \, \mathrm{d}x - \int_{0}^{R} \frac{x+2}{3 (x^2+x+1)} \, \mathrm{d}x \biggr]. \end{align*}

The last step follows from the fact that

$$ \text{PV}\!\! \int_{0}^{2} \frac{1}{x - 1} \, \mathrm{d}x = 0. $$

Now the last line can be computed by the usual calculus tricks, yielding

$$ \text{PV}\!\! \int_{0}^{\infty} \frac{1}{x^3 - 1} \, \mathrm{d}x = - \frac{\pi}{3\sqrt{3}}. $$

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If you really need to use contour integration, here is a trick using a keyhole contour. Let $\operatorname{Log}(z)$ be the branch of a complex logarithm chosen to satisfy $\operatorname{Arg}(z) \in [0, 2\pi)$. Its branch cut is the positive real axis. Now, consider the integral

$$ I_{\mathcal{C}}(s) = \int_{\mathcal{C}} f(z) \, \mathrm{d}z, \qquad f(z) = \frac{\exp(s \operatorname{Log}(z))}{z^3 - 1}, \tag{1} $$

where $-1 < s < 2$ and $\mathcal{C}$ is the keyhole contour shown below:

The shape of the contour $\mathcal{C}$ is determined by three parameters:

• $R$, the radius of the outer circular arc,

• $r$, the radius of the inner circular arcs, and

• $\epsilon$, the width of the keyhole.

By Cauchy integration formula, we have

\begin{align*} I_{\mathcal{C}}(s) &= 2\pi i \left( \frac{\exp(s \operatorname{Log}(z))}{3z^2} \biggr|_{z=e^{2\pi i/3}} + \frac{\exp(s \operatorname{Log}(z))}{3z^2} \biggr|_{z=e^{4\pi i/3}} \right) \\ &= \frac{2\pi i}{3} \left( e^{2\pi i (s-2) / 3} + e^{4\pi i (s-2) / 3} \right) \tag{2} \end{align*}

as long as $R > 1 > r > \epsilon$.

Next, we take suitable limits to $\text{(1)}$ so the integral can be related to OP's principal value. To begin with, we take the limit as $\epsilon \to 0^+$. This closes the opening of the keyhole, and the integral $\text{(1)}$ becomes:

\begin{align*} I_{\mathcal{C}}(s) &= \color{darkkhaki}{\int_{0}^{2\pi} f(Re^{i\theta}) iRe^{i\theta} \, \mathrm{d}\theta} - \color{#FF2B44}{\int_{0}^{2\pi} f(re^{i\theta}) ire^{i\theta} \, \mathrm{d}\theta} \\ &\quad - \color{green}{\int_{0}^{\pi} f(1+re^{i\theta}) ire^{i\theta} \, \mathrm{d}\theta} - \color{blue}{\int_{-\pi}^{0} f(1+re^{i\theta}) ire^{i\theta} \, \mathrm{d}\theta} \\ &\quad + \int_{r}^{1-r} f(x + i0^+) \, \mathrm{d}x + \int_{1+r}^{R} f(x + i0^+) \, \mathrm{d}x \\ &\quad - \int_{r}^{1-r} f(x - i0^+) \, \mathrm{d}x - \int_{1+r}^{R} f(x - i0^+) \, \mathrm{d}x \end{align*}

• Since $|f(Re^{i\theta})iRe^{i\theta}| \leq \frac{R^{s+1}}{R^3 - 1} $ for $R > 1$ and $s+1 < 3$, we get $$\color{darkkhaki}{\int_{0}^{2\pi} f(Re^{i\theta}) iRe^{i\theta} \, \mathrm{d}\theta} \to 0 \quad \text{as } R \to \infty. $$

• Since $|f(re^{i\theta})ire^{i\theta}| \leq \frac{r^{s+1}}{1 - r^3} $ for $0 < r < 1$ and $s+1 > 0$, we get $$ \color{#FF2B44}{\int_{0}^{2\pi} f(re^{i\theta}) ire^{i\theta} \, \mathrm{d}\theta} \to 0 \quad \text{as } r \to 0^+. $$

• We have \begin{align*} \color{green}{\int_{0}^{\pi} f(1+re^{i\theta}) ire^{i\theta} \, \mathrm{d}\theta} &= \int_{0}^{\pi} \frac{\exp(s \operatorname{Log}(1+re^{i\theta}))}{(1+re^{i\theta})^3 - 1} ire^{i\theta} \, \mathrm{d}\theta \\ &= i \int_{0}^{\pi} \frac{\exp(s \log \left|1+re^{i\theta}\right| + is \arctan(\frac{r\sin\theta}{1+r\cos\theta}))}{(1+re^{i\theta})^2 + (1+re^{i\theta}) + 1} \, \mathrm{d}\theta \\ &\to i \int_{0}^{\pi} \frac{\exp(s \operatorname{Log}(1))}{1^2 + 1 + 1} \, \mathrm{d}\theta \qquad \text{as } r \to 0^+ \\ &= \frac{i\pi}{3}. \end{align*}

• On the other hand, since $\operatorname{Im}(1+re^{i\theta}) < 0$ for $-\pi < \theta < 0$, our choice of branch cut gives \begin{align*} \color{blue}{\int_{-\pi}^{0} f(1+re^{i\theta}) ire^{i\theta} \, \mathrm{d}\theta} &= \int_{-\pi}^{0} \frac{\exp(s \operatorname{Log}(1+re^{i\theta}))}{(1+re^{i\theta})^3 - 1} ire^{i\theta} \, \mathrm{d}\theta \\ &= i \int_{-\pi}^{0} \frac{\exp(s \log \left|1+re^{i\theta}\right| + is [ \color{blue}{2\pi} + \arctan(\frac{r\sin\theta}{1+r\cos\theta}) ] )}{(1+re^{i\theta})^2 + (1+re^{i\theta}) + 1} \, \mathrm{d}\theta \\ &\to i \int_{-\pi}^{0} \frac{\exp(s \operatorname{Log}(1) + 2\pi i s)}{1^2 + 1 + 1} \, \mathrm{d}\theta \qquad \text{as } r \to 0^+ \\ &= \frac{i\pi e^{2\pi i s}}{3}. \end{align*}

• A similar reasoning as above shows that, for $x > 0$, we have $$ f(x + i0^+) = \frac{x^s}{x^3 - 1} \qquad\text{and}\qquad f(x - i0^+) = e^{2\pi i s} \frac{x^s}{x^3 - 1}. $$ Consequently, \begin{align*} &\int_{[r,1-r]\cup[1+r,R]} [f(x + i0^+) - f(x - i0^+)] \, \mathrm{d}x \\ &= (1 - e^{2\pi i s}) \int_{[r,1-r]\cup[1+r,R]} \frac{x^s}{x^3 - 1} \, \mathrm{d}x \\ &\to (1 - e^{2\pi i s}) \left( \text{PV}\!\! \int_{0}^{\infty} \frac{x^s}{x^3 - 1} \, \mathrm{d}x \right) \quad \text{as $r \to 0^+$ and $R \to \infty$}. \end{align*}

Combining altogether, it follows that

\begin{align*} I_{\mathcal{C}}(s) &= - \frac{i\pi}{3}(1 + e^{2\pi i s}) + (1 - e^{2\pi i s}) \left( \text{PV}\!\! \int_{0}^{\infty} \frac{x^s}{x^3 - 1} \, \mathrm{d}x \right). \tag{3} \end{align*}

Comparing $\text{(2)}$ with $\text{(3)}$, we get

\begin{align*} \text{PV}\!\! \int_{0}^{\infty} \frac{x^s}{x^3 - 1} \, \mathrm{d}x &= \frac{\pi i}{3} \cdot \frac{2( e^{2\pi i (s-2) / 3} + e^{4\pi i (s-2) / 3} ) + (1 + e^{2\pi i s})}{1 - e^{2\pi i s}} \\ &= - \frac{\pi}{3} \cdot \frac{2\cos(\pi (s-2) / 3) + \cos(\pi s)}{\sin(\pi s)} \tag{4} \end{align*}

whenever $s$ is not an integer. Finally, OP's integral can be obtained by letting $s \to 0$ to $\text{(4)}$, yielding

$$ \text{PV}\!\! \int_{0}^{\infty} \frac{1}{x^3 - 1} \, \mathrm{d}x = - \frac{\pi}{3\sqrt{3}}. $$