Calculate the integral using complex analysis $$\int\limits_{0}^{\infty }\frac{\ln x}{\sqrt[3]{x}(x+8)}dx$$

This integral can easily be calculated using integration by parts, but I was unable to solve it using residues... I don't know which pole should be and which direction

Could you please tell me how to solve this integral using residues?

  • 1
    $\begingroup$ What do you mean by using deductions? $\endgroup$
    – Mark
    Nov 19, 2023 at 0:23
  • $\begingroup$ en.wikipedia.org/wiki/Residue_(complex_analysis) $\endgroup$
    – Dmitry
    Nov 19, 2023 at 0:25
  • $\begingroup$ Well, that's called using residues. The word "deductions" doesn't even appear in the Wikipedia page. $\endgroup$
    – Mark
    Nov 19, 2023 at 0:26
  • $\begingroup$ If you want to solve it with residues, try to use the so called keyhole contour, or "Pacman" contour. Are you familiar with it? $\endgroup$
    – Mark
    Nov 19, 2023 at 1:18
  • $\begingroup$ No, I'm not familiar with "Pacman" $\endgroup$
    – Dmitry
    Nov 19, 2023 at 1:21

3 Answers 3


I used Feymann's trick: $$\text{Let}\quad f\left(x\right)=\int_{0}^{\infty}\frac{t^{x}}{t+8}dt\quad\Rightarrow\quad f'\left(-\frac{1}{3}\right)=\int_{0}^{\infty}\frac{\ln\left(t\right)}{\sqrt[3]{t}\left(t+8\right)}dt$$ $$f(x)=-8^x\pi\csc(\pi x)\quad\Rightarrow\quad f'\left(-\frac{1}{3}\right)=\frac{\pi}{3} (\pi + \sqrt{3} \ln(8))$$

I don't know what you mean by "deductions" but this is a much faster method than integration by parts

  • $\begingroup$ I mixed up the words. That's what I meant. Could you show the second way through the Residue? en.wikipedia.org/wiki/Residue_(complex_analysis) $\endgroup$
    – Dmitry
    Nov 19, 2023 at 0:48

We seek to evaluate

$$J = \int_0^\infty \frac{x^{-1/3}}{x+8} \log x \; dx.$$

We introduce the function

$$F(z) = \frac{\exp(-1/3\times \mathrm{Log}(z))}{z+8} \mathrm{Log}(z)$$

where the logarithm has the branch cut on the positive real axis and $0\le \mathrm{arg} \; \mathrm{Log} \lt 2\pi.$

We now use a keyhole contour to integrate $F(z)$ with the slot on the positive real axis (note that the pole at $z=-8$ is not on the cut) and the large circle of radius $R$ while the circle around the origin has radius $\varepsilon$. We have on the large circle $\Gamma_1$ (number segments in counterclockwise fashion)

$$\frac{\exp(-1/3 \times \mathrm{Log}(R \exp(i\theta)))} {R\exp(i\theta) + 8} \mathrm{Log}(R\exp(i\theta)).$$

Computing the norm we find $2\pi R \times \frac{R^{-1/3}}{R} \log R \rightarrow 0$ as $R\rightarrow\infty.$ (Remember that $|\exp(-1/3\times i\theta)|=1.$)

Similarly for the small circle $\Gamma_3$ we have

$$\frac{\exp(-1/3 \times \mathrm{Log}(\varepsilon \exp(i\theta)))} {\varepsilon \exp(i\theta) + 8} \mathrm{Log}(\varepsilon\exp(i\theta)).$$

Compute the norm once more to get $2\pi\varepsilon \times \frac{\varepsilon^{-1/3}}{8} \log\varepsilon \rightarrow 0$ as $\varepsilon\rightarrow 0.$

Now we get the integral $J$ on the line $\Gamma_0$ just above the slot and below the slot on $\Gamma_2$ we find

$$\int_{\Gamma_2} \frac{\exp(-1/3\times \log (z) - 1/3 \times 2\pi i )}{z+8} (\log(z) + 2\pi i) \; dz \\ = - \exp(-2\pi i/3) J - 2\pi i \exp(-2\pi i/3) \int_0^\infty \frac{x^{-1/3}}{x+8} \; dx \\ = - \exp(-2\pi i/3) J - 2\pi i \exp(-2\pi i/3) K.$$

Now to compute $K$ we use the same keyhole contour and get for the circular bounds $2\pi R \frac{R^{-1/3}}{R} \rightarrow 0$ as $R\rightarrow\infty$ as well as $2\pi\varepsilon \frac{1}{8} \frac{1}{\varepsilon^{1/3}} \rightarrow 0$ as $\varepsilon\rightarrow 0.$ It follows by the Cauchy Residue Theorem that

$$K (1-\exp(-2\pi i/3)) = 2\pi i \; \underset{z=-8}{\mathrm{res}} \; \frac{z^{-1/3}}{z+8} = 2\pi i \times (-8)^{-1/3} \\ = 2\pi i \times \exp(-1/3 \times \mathrm{Log}(-8)) \\ = 2\pi i \times \frac{1}{2} \times \exp( -1/3 \times \pi i) \\ = \pi i \exp( -1/3 \times \pi i).$$

We obtain for $K$

$$K = \pi i \frac{\exp(-\pi i/3)} {1-\exp(-2\pi i/3)} \\ = \pi i \frac{1}{\exp(\pi i/3) - \exp(-\pi i/3)} = \frac{\pi}{2 \sin(\pi/3)} = \frac{\sqrt{3}}{3} \pi.$$

We thus obtain for $J$

$$J (1-\exp(-2\pi i/3)) - 2\pi i \times \exp(-2\pi i/3) \pi \frac{\sqrt{3}}{3} \\ = 2\pi i \times \; \underset{z=-8}{\mathrm{res}} \; \frac{z^{-1/3}}{z+8} \;\mathrm{Log}(z).$$

The residue works out to

$$\frac{1}{2} \exp(-1/3\times \pi i) (3\log(2) + \pi i).$$


$$J (1-\exp(-2\pi i/3)) \\ = 2\pi i \times \exp(-2\pi i/3) \pi \frac{\sqrt{3}}{3} + \pi i \exp(-\pi i/3) (3\log 2 + \pi i).$$


$$J (\exp(\pi i/3)-\exp(-\pi i/3)) \\ = 2\pi i \times \exp(-\pi i/3) \pi \frac{\sqrt{3}}{3} + \pi i (3\log 2 + \pi i)$$


$$J \sin(\pi/3) = \pi \times \exp(-\pi i/3) \pi \frac{\sqrt{3}}{3} + \frac{1}{2} \pi (3\log 2 + \pi i).$$

Now we know $J$ must be a real number so we may omit the imaginary components on the right to get

$$J \frac{\sqrt{3}}{2} = \pi \times \cos(\pi/3) \pi \frac{\sqrt{3}}{3} + \frac{3}{2} \pi \log 2.$$

We have at last

$$J = \pi \times \pi \frac{1}{3} + \frac{3}{\sqrt{3}} \pi \log 2$$

which is

$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{3} \pi^2 + \pi \sqrt{3} \log 2.}$$


First, write

$$I = \int_0^\infty \frac{\log x}{x^{1/3} (x+8)} \, dx \stackrel{x\mapsto x^3}= 9 \int_0^\infty \frac{x \log x}{x^3+8} \, dx$$

Now consider the integral in $\Bbb C$,

$$J = \oint_\Gamma \frac{z \log^{\color{red}2}z}{z^3+8} \, dz$$

where $\Gamma$ is a keyhole contour like the one pictured in Wikipedia's Example 4, taking the branch cut along the positive real axis. The integrals along the larger and smaller circular arcs will vanish as their respective radii approach $\infty$ and $0$. As we approach the real axis from above, we have on the upper line segment that $\log z\to\log|z|$; from below, on the lower line segment, $\log z\to\log|z|+i2\pi$.

By the residue theorem,

$$J = i2\pi \sum \underset{z=\omega}{\operatorname{Res}} \frac{z\log^2z}{z^3+8} = \frac{4\pi^3}{3\sqrt3} - i \left(\frac{4\pi^3}{27} + \frac{4\pi^2}{3\sqrt3} \log2\right)$$

where $\omega^3=-8$.

Now, parameterizing the upper/lower line segments by $z=x\pm i\varepsilon$, respectively, we have

$$\begin{align*} J &= \lim_{R\to\infty\\\varepsilon\to0^+} \int_\varepsilon^R \frac{(x+i\varepsilon) \log^2(x+i\varepsilon)}{(x+i\varepsilon)^3+8} \, dx + \int_R^\varepsilon \frac{(x-i\varepsilon) \left(\log(x-i\varepsilon) + i2\pi\right)^2}{(x-i\varepsilon)^3+8} \, dx \\ &= \int_0^\infty \frac{x \log^2x}{x^3+8} \, dx - \int_0^\infty \frac{x \left(\log^2x + i4\pi \log x - 4\pi^2\right)}{x^3+8} \, dx \\ &= -i4\pi \int_0^\infty \frac{x \log x}{x^3+8} \, dx + 4\pi^2 \int_0^\infty \frac{x}{x^3+8} \, dx \end{align*}$$

so that upon matching up the imaginary parts, we conclude

$$\frac19 I = \frac{\pi^2}{27} + \frac{\pi}{3\sqrt3}\log2$$


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