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$$I=\int_0^2 x(8-x^3)^{\frac{1}{3}} \, dx$$ Being inept with integrals, this is my try,

$$\int x(8-x^3)^{\frac{1}{3}} \, dx = (-1)\int x(x^3-8)^{\frac{1}{3}} \, dx $$

Using the substitution method (Chebyshev); for $\int x^m(a+bx^n)^p \,dx$, where $m, n, p ∈ Q$, If $\frac{m+1}{n}+p ∈ Z$, then substitute $(ax^{-n}+b)=t^K$, where $K$ is the denominator of $p$;

$$\frac{(x^3-8)^{\frac{1}{3}}}{x}=t ,\, dx=\frac{dt}{\left(\frac{x}{(x^3-8)^{\frac{2}{3}}}-\frac{(x^3-8)^{\frac{1}{3}}}{x^2}\right)}$$

$$\int \frac{8t^3}{(t-1)^2(t^2+t+1)^2} \, dt=8 \left( \frac{1}{3} \int \frac{1}{t^3-1}\,dt-\frac{1}{3} \int \frac{t}{t^3-1} \, dt\right)$$

Continuing the partial fractions, the final result is; $$\frac{8}{3} \left( \frac{\ln(t-1)}{3}-\frac{t}{t^3-1}-\frac{\arctan \left(\frac{2t+1}{\sqrt3}\right)}{\sqrt3} \right)-\frac{4}{9} \ln(t^2+t+1) $$

On putting the bounds, $I=\frac{16\pi}{(\sqrt3)^5}$

However, this is a very tedious method. In search for alternate method, i tried for beta functions, it resembled the format (almost). To bring it into the beta function form;

$B(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1} \, dx$;

I tried with $x^3=8y$, which gave $I=\frac{8}{3} B\left(\frac{2}{3},\frac{4}{3}\right)$ and using up almost all the properties;

  1. $B(p,q)=\frac{\Gamma_p \Gamma_q}{\Gamma_{p+q}}$

  2. $\Gamma_{p+1}=p(\Gamma_p)$

  3. $\Gamma_p \Gamma_{1-p}=\frac{\pi}{sinp\pi}$, you get $I=\frac{16\pi}{(\sqrt3)^5}$.

I am interested in learning more ways to evaluate this integral.

xpaul
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NadiKeUssPar
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5 Answers5

5

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{Res}}\newcommand{\D}{\mathfrak{D}}$Take $x\mapsto x^{1/3}$ in the integral to obtain the equivalent expression: $$I=\frac{1}{3}\int_0^8x^{-1/3}(8-x)^{1/3}\d x$$Define a logarithm $\log':\Bbb C^\star\to\Bbb C$ through $-\pi\le\arg<\pi$ and a logarithm $\log:\Bbb C^\star\to\Bbb C$ through $0\le\arg<2\pi$. For $z\in\Bbb C^\star$, denote by $z^{-1/3}$ the quantity $\exp\frac{-1}{3}\log(z)$ and for $z\in\Bbb C\setminus\{8\}$ denote by $(8-z)^{1/3}$ the quantity $\exp\frac{1}{3}\log'(8-z)$.

Define $f:\Bbb C\setminus[0,8]\to\Bbb C$, $z\mapsto\frac{1}{3}z^{-1/3}(8-z)^{1/3}$. It follows from the choices of logarithm that the branch cuts on $[0,\infty),[8,\infty)$ cancel nicely to ensure $f$ is in fact holomorphic.

By considering the asymptotics of $f$ as $z\to\infty$ along the positive real axis (for simplicity's sake, and it's valid to restrict attention to that case since the power series exists), it's not too hard to see the $f$'s power series at $\infty$ must take the following form: $$f(z)=\frac{1}{3}e^{-\pi i/3}-\frac{8}{9}e^{-\pi i/3}z^{-1}+O(z^{-2})$$It follows that: $$\oint_{\D}f(z)\d z=2\pi i\res(f;\infty)=\frac{16\pi i}{9}e^{-\pi i/3}$$Where $\D$ is a clockwise dogbone contour that runs $0+i\delta\to8+i\delta\to8-i\delta\to-i\delta\to i\delta$ for some $\delta>0$. The section $-i\delta\to i\delta$ is a clockwise semicircular contour of radius $\delta$ that passes to the left of zero, and the section $8+i\delta\to8-i\delta$ is a clockwise semicircular contour of radius $\delta$ that passes to the right of $8$.

Letting $\delta\to0^+$ concludes: $$(1-e^{-2\pi i/3})I=\frac{16\pi i}{9}e^{-\pi i/3}$$Noting that $f(z)\to e^{-2\pi i/3}g(x)$ as $z\to x$ from below the real axis, if $g$ denotes the integrand of $I$, and noting also that the small semicircular integrals vanish since $\lim_{z\to0}zf(z)=0$ and $\lim_{z\to8}(z-8)f(z)=0$.

Thus: $$I=\frac{16\pi i}{9}\cdot\frac{\csc(\pi/3)}{2i}=\frac{16\pi}{9\sqrt{3}}$$As desired.

FShrike
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    This was quite interesting, thanks. – NadiKeUssPar Jul 04 '23 at 11:51
  • FShrike, $\mathfrak D$ contains $0$ and $8$ but the function $f(z)$ is not defined at that two points, moreover I cannot understand why you consider $\text{Res}(f;\infty)$ if $\infty$ is external to $\mathfrak D$. And why $f(z)\to e^{-2\pi i/3}g(x)$ as $z\to x$ from below the real axis ? It seems that $f(z)\to g(x)$. I am sorry but I cannot understand a lot of things about your answer. And what do you mean when you write “the branch cuts on $[0,\infty),[8,\infty)$ cancel nicely to ensure $f$ is in fact holomorphic” ? Please could you explain all it with more details ? – Angelo Jul 04 '23 at 13:25
  • @Angelo Note for any $\delta>0$, $f$ is defined at every single point on the contour $\frak{D}$. So the integration is well-defined. The residue at infinity is always to be considered, and for a clockwise contour $\infty$ is internal to the contour on the Riemann sphere. That’s just a nice visualisation though, I know no advanced complex theory; it is nevertheless an extension of the residue theorem to include the residue at $\infty$, proven by taking a Laurent series expansion of $f$ at infinity. – FShrike Jul 04 '23 at 13:33
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    Note that $\log z\to\ln x+2\pi i$ as $z\to x>0$ from below the real axis. Therefore $z^{-1/3}\to\exp(\ln x+2\pi i)/3=x^{-1/3}\cdot e^{-2\pi i/3}$, so overall $f(z)\to e^{-2\pi i/3}g(x)$ – FShrike Jul 04 '23 at 13:34
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    Finally for why $f$ is holomorphic outside of $[0,8]$: the only possible problem could be on $(8,\infty)$ where the exponentials individually have discontinuities. However if you take limits from above and below you find that the product $f$ doesn’t have any discontinuities; the ‘jumps’ by $e^{2\pi i/3}$ from each exponential will literally cancel. Think about why the logarithm is not holomorphic on the whole plane, if you don’t know what I mean in the last two comments – FShrike Jul 04 '23 at 13:36
  • @Angelo "On $\mathfrak{D}$" is potentially ambiguous. The contour $\mathfrak{D}$ itself is just the line; to integrate $\oint_{\mathfrak{D}}f(z),\mathrm{d}z$, I need only that for every $t\in[0,1]$ that $f(\gamma(t))$ is defined, where $\gamma$ is some parametrisation of $\mathfrak{D}$ (... and that $f$ is integrable). Although $[0,8]$ is "enclosed" by $\mathfrak{D}$, $[0,8]$ does not intersect (the trace of) $\mathfrak{D}$. It is like integrating $\int_{[0,1]\cup[2,3]}h(x),\mathrm{d}x$, the values of $h$ on the "enclosed" interval $(1,2)$ are quite irrelevant – FShrike Jul 04 '23 at 15:04
  • @FShrike, $0+i\delta\to8+i\delta\to8-i\delta\to-i\delta\to i\delta$ contains $0$ and $8$. Why do you say that $f$ is defined on $\mathfrak D$? $0\in -i\delta\to i\delta$ and $8\in 8+i\delta\to 8-i\delta$. Am I wrong ? Moreover, I believed that we have to consider only the poles inside the contour $\mathfrak D$ and for a clockwise contour we get the opposite result. Moreover, you said that $-\pi\leqslant\arg(\log z)<\pi$, but now you write that $\log z\to\ln x+2\pi i$ as $z\to x$ from below the real axis. Why ? – Angelo Jul 04 '23 at 15:04
  • Ah sorry, I now see your confusion. – FShrike Jul 04 '23 at 15:04
  • As is implied by "dogbone contour" (a known term) (but I will now edit and clarify this) the sections $-i\delta\to i\delta$ and $8+i\delta\to8-i\delta$ are semicircular indents that do not cross $[0,8]$ – FShrike Jul 04 '23 at 15:05
  • I made a typo with $\log z\to\ln x+2\pi i$, I meant $\log'z\to\ln x+2\pi i$. I've now swapped the meaning of $\log$ and $\log'$ in the post so as to correct this typo (I can't edit the comment). So, now it is correct that $\log z\to\ln x+2\pi i$. – FShrike Jul 04 '23 at 15:06
  • Regarding "consider only the poles inside the contour" I can't really explain this without just writing a whole proof of this extension of the residue theorem. The usual residue theorem works by expanding in Laurent series about each pole inside the contour, here I am using a version where I expand in Laurent series about each pole outside the contour. Either is fine, both are correct theorems and yield the same answer if both apply. Here, the usual theorem doesn't apply because $f$ can't be extended to a meromorphic function on the interior of $\mathfrak{D}$. – FShrike Jul 04 '23 at 15:13
  • You can think of it as just using the usual theorem on the modified function $g(z)=f(1/z)$ which is holomorphic on $B(0,1/8)$ with an essential singularity at $z=0$ – FShrike Jul 04 '23 at 15:15
  • @FShrike, thank you for you explanation. Congratulations for your answer (+1). – Angelo Jul 04 '23 at 15:46
  • @Angelo thanks, and no worries. Dogbone integration is my favourite trick – FShrike Jul 04 '23 at 16:55
  • @FShrike, how have you obtained the $f(z)$’s power series at $\infty$ ? I have obtained it in the following way: – Angelo Jul 04 '23 at 20:57
  • @FShrike, $$\begin{align}f(z)&=\frac13\exp\left(-\frac13\log z+\frac13\log’(8-z)\right)=\[3pt]&=\frac13\exp\left(\frac13\log’\left(\tfrac8z-1\right)\right)=\[3pt]&=\frac13\exp\left[\frac13\bigg(!!-i\pi+\log’\left(1-\tfrac8z\right)\bigg)\right]=\[3pt]&=\frac13\exp\left[\frac13\left(-i\pi-\frac8z+o!\left(!z^{-1}!\right)\right)\right]=\[3pt]&=\frac13\exp\left(-\frac{i\pi}3\right)\cdot\exp\left(-\frac8{3z}+o!\left(!z^{-1}!\right)\right)=\[3pt]&=\frac13e^{-\tfrac{i\pi}3}\left(1-\frac8{3z}+o!\left(!z^{-1}!\right)\right)\end{align}$$ Is it correct my way to get it ? Please correct it. – Angelo Jul 04 '23 at 20:57
  • @FShrike, tell me how you have obtained it, please. What is your way to get the $f(z)$’s power series at $\infty$ ? – Angelo Jul 04 '23 at 20:58
  • @Angelo That's all correct if you assume $z$ is a large real number. Then this result extends, because $f$ is holomorphic, to being valid for all large $z$. I should also note the residue at infinity is minus the $[z^{-1}]$ coefficient – FShrike Jul 04 '23 at 21:08
  • @FShrike, thank you for your answer, but please could tell me how you obtained the power series of $f(z)$ at infinity ? – Angelo Jul 05 '23 at 02:42
  • @Angelo Basically the same way you did! As I said, that’s correct – FShrike Jul 05 '23 at 09:43
5

Integrate the following general form by parts

\begin{align} &\int_0^a x(a^3-x^3)^{1/3} \, dx\\ =& \int_0^a \frac{(a^3-x^3)^{1/3}}x\, d(\frac{x^3}3) = \frac{a^3}3\int_0^a\frac x{(a^3-x^3)^{2/3}}\overset{a^3-x^3\to x^3}{dx}\\ =&\ \frac{a^3}3\int_0^a\frac 1{\sqrt[3]{a^3-x^3}}{dx} = \frac{a^3}3\cdot \frac{2\pi}{3\sqrt3}=\frac{2\pi a^3}{9\sqrt3} \end{align} where $ \int_0^a\frac {dx}{\sqrt[3]{a^3-x^3}}= \frac{2\pi}{3\sqrt3}$.

Quanto
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4

Let $\displaystyle x = \frac{2}{(1+t)^{1/3}}$. Then

$$ I:= \int_{0}^{2}x\left(8-x^{3}\right)^{\frac{1}{3}}dx = \frac{8}{3}\int_{0}^{\infty}\frac{t^{\frac{1}{3}}}{\left(t+1\right)^{2}}dt = \frac{8}{3}\cdot\frac{\left(1-\frac{4}{3}\right)\pi}{\sin\left(\frac{4}{3}\pi\right)} = \frac{16\pi}{9\sqrt{3}}. $$

Where for evaluating that last integral, we use this.

Alternatively, let $x = \displaystyle \frac{2}{\left(u^{3}+1\right)^{1/3}}$. Then

$$ \begin{align} I &:= \int_{0}^{2}x\left(8-x^{3}\right)^{\frac{1}{3}}dx \\ &= 8\int_{0}^{\infty}\frac{u^{3}}{\left(1+u^{3}\right)^{2}}du \\ &= 8\int_{0}^{\infty}\frac{u^{3}+1-1}{\left(1+u^{3}\right)^{2}}du \\ &= 8\int_{0}^{\infty}\frac{1}{1+u^{3}}du-8\int_{0}^{\infty}\frac{1}{\left(1+u^{3}\right)^{2}}du \\ &= 8\int_{0}^{\infty}\frac{1}{1+u^{3}}du-8\left(\left[\frac{u}{3\left(u^{3}+1\right)}\right]_{0}^{\infty}+\frac{2}{3}\int_{0}^{\infty}\frac{1}{u^{3}+1}du\right) \tag{1}\\ &= 8\int_{0}^{\infty}\frac{1}{1+u^{3}}du-\frac{16}{3}\int_{0}^{\infty}\frac{1}{u^{3}+1}du \\ &= \frac{8}{3}\int_{0}^{\infty}\frac{1}{1+u^{3}}du \\ &= \frac{16\pi}{9\sqrt{3}}. \tag{2}\\ \end{align} $$

$(1)$ We use Ostrogradsky's method by letting $P(u) = 1$, $Q(u) = (1+u^3)^2$, $Q_1(u) = Q_2(u) = u^3+1$, and we find $P_1(u) = u/3$ and $P_2 (u) = 2/3$.

$(2)$ We use this.

Accelerator
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    Oh @Accelerator! I literally found its closed form like this. Cheers, What I think is while trying to evaluate integrals feynman's technique, beta and gamma functions can be very helpful at any time. – Lucky Chouhan Jul 04 '23 at 11:51
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    @LuckyChouhan You might want to post that as an answer then. :) – Accelerator Jul 04 '23 at 11:51
  • You got it. But you already posted so I don't think I should. – Lucky Chouhan Jul 04 '23 at 11:53
  • Great answer, would love to share the result you linked at your last line, for $\int_0^{\infty} \frac{1}{1+x^n} dx = \frac{\pi}{n}cosec(\frac{\pi}{n})$, n being natural – NadiKeUssPar Jul 04 '23 at 11:56
  • @Ercanayan Honestly, I was too lazy to provide my own work for that last line. When I got to that last line, I figured it was shared a billion times on the Internet so I decided to link one result. – Accelerator Jul 04 '23 at 20:48
2

The substitution $x=2\sin^{\frac 23}\theta$ transforms the integral into $$ \begin{aligned} I & =\frac{16}{3} \int_0^{\frac{\pi}{2}} \sin ^{\frac{1}{3}} \theta \cos ^{\frac{5}{3}} \theta d \theta \\ & =\frac{16}{3} \cdot\frac{1}{2} B\left(\frac{1}{3}, \frac{4}{3}\right) \\ & =\frac{8 \pi}{3} \cdot \csc \frac{\pi}{3} \quad (\textrm{By } B(x,1-x)=\pi\csc (\pi x))\\ & =\frac{16 \pi}{3 \sqrt{3}} \end{aligned} $$

Lai
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2

Here is a variation based upon binomial expansion and hypergeometric functions. We also use the rising factorial notation $(a)_{n}:=a(a+1)\cdots(a+n-1)$.

We obtain \begin{align*} \color{blue}{\int_{0}^{2}}\color{blue}{x\left(8-x^3\right)^{1/3}\,dx} &=2\int_{0}^{2}x\sum_{n=0}^{\infty}\binom{\frac{1}{3}}{n}\left(-\frac{x^3}{8}\right)^n\,dx\\ &=2\sum_{n=0}^{\infty}\binom{\frac{1}{3}}{n}\left(-\frac{1}{8}\right)^n\int_{0}^2x^{3n+1}\,dx\\ &=2\sum_{n=0}^{\infty}\binom{\frac{1}{3}}{n}\left(-\frac{1}{8}\right)^n\left.\frac{1}{3n+2}x^{3n+2}\right|_0^2\\ &=8\sum_{n=0}^{\infty}\underbrace{\binom{\frac{1}{3}}{n}\frac{(-1)^n}{3n+2}}_{:=t_n}\\ &=8\sum_{n=0}^{\infty}t_n=8t_0\sum_{n=0}^{\infty}\prod_{j=0}^{n-1}\frac{t_{j+1}}{t_j}\\ &=4\sum_{n=0}^{\infty}\prod_{j=0}^{n-1}\binom{\frac{1}{3}}{j+1}\frac{(-1)^{j+1}}{3j+5} \binom{\frac{1}{3}}{j}^{-1}(-1)^{-j}(3j+2)\\ &=4\sum_{n=0}^{\infty}\prod_{j=0}^{n-1}\frac{\left(j-\frac{1}{3}\right)\left(j+\frac{2}{3}\right)} {(j+1)\left(j+\frac{5}{3}\right)}\\ &=4\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{3}\right)_n\left(\frac{2}{3}\right)_n}{\left(\frac{5}{3}\right)_n}\,\frac{1}{n!}\\ &=4{}_2F_{1}\left(-\frac{1}{3},\frac{2}{3};\frac{5}{3};1\right)\tag{1}\\ &=4\frac{\Gamma\left(\frac{5}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma(2)\Gamma(1)}\tag{2}\\ &=4\frac{2\pi}{9\sqrt{3}}\Gamma(3)\tag{3}\\ &\,\,\color{blue}{=\frac{16\pi}{9\sqrt{3}}} \end{align*} according to the claim.

Comment:

  • In (1) we write the sum as hypergeometric $_2F_1$ function evaluated at $z=1$ .

  • In (2) we recall a theorem from C. F. Gauss [1812] (see e.g. Theorem 2.2.2 in Special Functions by G.E. Andrews, R. Askey and R. Roy) which is \begin{align*} {_2F_1}(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\tag{4} \end{align*} if $\Re(c-a-b)>0$. We derive from (1) $\Re\left(\frac{5}{3}+\frac{1}{3}-\frac{2}{3}\right)=\frac{4}{3}>0$ and we can apply (4).

  • In (3) we use $\Gamma(2)=\Gamma(1)=1$ and the Gamma triplication formula \begin{align*} \Gamma (3z)=(2 \pi)^{-1}3^{3z-1/2}\Gamma(z)\Gamma \left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right) \end{align*} evaluated at $z=1$.

Markus Scheuer
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