3
$\begingroup$

Whilst looking at someone's vector calculus problem, they mentioned that, making use of Green's Theorem, they had to express the line integral of the boundary of $x^{10}+y^{10}\leq 1$ in terms of its area. The thing is they gave you the area as computed with Mathematica to be $4\Gamma^2(11/10)/\Gamma(6/5)\approx 3.943.$

And I was wondering how you'd prove this: $$\mathcal{A}=\iint_{x^{10}+y^{10}\leq1}dydx=2\int_{-1}^1\sqrt[10]{1-x^{10}}dx=\dfrac{4\Gamma^2(11/10)}{\Gamma(6/5)}.$$

$\endgroup$
5
  • 1
    $\begingroup$ It is a method due to Dirichlet; it is in Whittaker and Watson, A Course of Modern Analysis. First edition was 1902, on pages 191-193. I found an online scan ... let me see if I can find it again. Well, this is the third edition, ia801601.us.archive.org/6/items/courseofmodernan00whit/… $\endgroup$
    – Will Jagy
    Jul 27, 2023 at 21:41
  • 1
    $\begingroup$ page n8mbered 258, pdf page called 272 $\endgroup$
    – Will Jagy
    Jul 27, 2023 at 21:50
  • $\begingroup$ Are you trying to do this with Green’s Theorem? Do the line integral $$\frac12\int_C -y\,dx+x\,dy,$$ parametrizing $C$ by $x=(\cos t)^{1/5}$, $y=(\sin t)^{1/5}$. $\endgroup$ Jul 27, 2023 at 23:50
  • $\begingroup$ We could also apply what you said: Choosing $\mathbf{F}=x\boldsymbol{j}$, we get that $$\iint_A (\partial_x F_y-\partial_y F_x)dydx=\iint_A dydx,$$ and by Green's theorem we know that is also equal to $$\oint_{\partial A}\mathbf{F\bullet dr}=\oint_{\partial A}xdy$$ and by parametrizing $x$ and $y$ like you said, we get $$\frac{1}{5}\int^{2\pi}_0 \cos^{\frac{6}{5}}t\sin^{-\frac{4}{5}}tdt,$$ which has somewhat the form of the beta function: $$\mathfrak{B}(x,y)=2\int_0^{\frac{\pi}{2}}\cos^{2x-1}\phi\sin^{2y-1}\phi d\phi.$$ And from here IDK how to proceed. $\endgroup$
    – Conreu
    Jul 28, 2023 at 0:39
  • $\begingroup$ I am not sure why but when computing the integral numerically, I don't get what I'm supposed to... $\endgroup$
    – Conreu
    Jul 28, 2023 at 1:11

2 Answers 2

5
$\begingroup$

We can tackle a more general question: that of the unit $\ell^p$ ball in $\mathbb{R}^n$, i.e. the volume of the body given by $$ \{ (x_1,\cdots,x_n) \in \mathbb{R}^n \mid |x_1|^p + |x_2|^p + \cdots + |x_n|^p \le 1 \} $$ Note that your case is $n=2$ and $p=10$.

Let $V_{n,p}$ denote the volume of this ball. Then, analogous to your transformation from a multi-integral to a single integral, we see $$ V_{n,p} = 2 V_{n-1,p} \int_{0}^1 (1-x^p)^{(n-1)/p} \, \mathrm{d}x $$ To see this, note that $$ |x_1|^p + |x_2|^p + \cdots + |x_n|^p \le 1 \iff |x_2|^p + \cdots + |x_n|^p \le 1 - |x_1|^p $$ and rewrite as $$ V_{n,p} = \int_{-1}^1 \mathrm{d}x_1 \int_{ \sum_2^n |x_i|^p \le 1 - |x_1|^p} \mathrm{d} x_2 \cdots \mathrm{d} x_n $$ with the right integral as the volume of an $(n-1)$-dimensional ball.

Now, with the substitution of $u=x^p$, we see that $$ x=u^{1/p} \qquad \mathrm{d}u = px^{p-1} \, \mathrm{d} x \qquad \mathrm{d} x = \frac 1 p u^{\frac 1 p - 1} $$ and hence $$ \int_{0}^1 (1-x^p)^{(n-1)/p} \, \mathrm{d}x = \frac 1 p \int_{0}^1 u^{\frac 1 p - 1} (1-u)^{(n-1)/p} \, \mathrm{d}u $$ The beta function is defined by $$ B(x,y) := \int_0^1 t^{x-1} (1-t)^{y-1} \, \mathrm{d} t \equiv \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $$ so we hence see $$ V_{n,p} = \frac 2 p V_{n-1,p} \cdot \frac{ \Gamma(\frac 1 p ) \Gamma( \frac{n-1}{p} + 1 )}{\Gamma( \frac n p + 1 )} $$ Apply the recursion property of the gamma function, $\Gamma(z+1) = z \Gamma(z)$, to then get $$ V_{n,p} = 2 V_{n-1,p} \cdot \frac{ \Gamma(\frac 1 p + 1 ) \Gamma( \frac{n-1}{p} + 1 )}{\Gamma( \frac n p + 1 )} $$ This gives us a recursive relation; iterating backwards to $V_{1,p} = 2$, we then get the desired result of $$ V_{n,p} = 2^n \frac{\Gamma^n(\frac 1 p + 1)}{\Gamma(\frac n p + 1)} $$ With $n=2$ and $p=10$, we see your desired result: $$ V_{2,10} = 4 \frac{\Gamma^2(11/10)}{\Gamma(6/5)} $$

$\endgroup$
1
  • $\begingroup$ Very complete solution on your part: thank you very much. $\endgroup$
    – Conreu
    Jul 27, 2023 at 22:14
3
$\begingroup$

This is a corollary to the relationship between the gamma and beta functions:$$ \mathrm{B}(x,y) \doteq \int_0^1 t^{x-1}(1-t)^{y-1} dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}. $$ To see this we use the substitution $u = t^n$ to find that $$ \int_0^1 (1-t^n)^{1/n}\,dt = \int_0^1 (1-u)^{1/n} u^{-1+1/n}\,du = \frac{1}{n}\mathrm{B}(1+1/n,1/n) = \frac{\Gamma(1+1/n)\Gamma(1/n)}{n \Gamma(1+2/n)} = \frac{\Gamma(1+1/n)^2}{\Gamma(1+2/n)}. $$ The final step uses the identity $\Gamma(x+1) = x \Gamma(x)$. Interestingly, if we expand this formula in powers of $1/n$ we find out how fast the area converges to that of the $2 \times 2$ square: $$ 4 \frac{\Gamma(1+1/n)^2}{\Gamma(1+2/n)} = 4 - \frac{2\pi^2}{3n^2} + O(n^{-3}).$$

$\endgroup$
1
  • $\begingroup$ Really interesting and straightforward, thank you. $\endgroup$
    – Conreu
    Jul 27, 2023 at 22:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .