Two Equations, Two Unknowns and the Beta Function

I spotted this little problem on Twitter (see Figure 1):

Figure 1

The proximity of $+31$ to $+32$ should alert us to the possibility that either $x=+2$ or $y=+2$ is a solution. After that, the corresponding $x$ and $y$ values fall into place. However, let's suppose that we didn't spot this and I have to confess I didn't. In that case, we follow the conventional method. 

At first sight, it looks like you will end up with a quintic equation after substituting $y=1-x$ into $x^5+y^5=31$ and there is no general formula for solving quintic equations. However, the $x^5$ terms cancels out nicely leaving us with a quartic equation. Let's see how:$$\begin{array}{rl}{x}^5+(1-x{)}^5& =31\\ x^5+(1-5x+10x^2-10x^3+5x^4-x^5)& =31\\ 5x^4-10x^3+10x^2-5x-30& =0\\ 5x^3(x-2)+5(2x^2-x-6)& =0\\ 5x^3(x-2)+5(2x+3)(x-2)& =0\\ (x-2)(5x^3+10x+15)& =0\\ 5(x-2)(x^3+2x+3)& =0\\ 5(x-2)(x+1)(x^2-x+3)& =0\end{array}$$Thus $x=2$ and $y=-1$ or $x=-1$ and $y=2$. Using GeoGebra, this can be confirmed visually as shown in Figure 2:

Figure 2

Differentiation reveals a little more about the graph of $x^5+y^5=31$:$$\begin{array}{rl}{y}^{\prime }& =\frac{-x^4}{y^4}\\ & =-\frac{x^4}{\sqrt[5]{31-x^5}}\end{array}$$It can be seen that $y^{\prime }$ is undefined when $x=\sqrt[5]{31}\approx 1.9873$ and thus the graph is vertical at this point. If we zoom in a little, this becomes clearer (see Figure 3). There is a point of horizontal inflection when $x=0$.

Figure 3

This was a useful little exercise as I got to revisit the binomial expansion for $(a+b{)}^5$ and the factor theorem in order to factorise $x^3+2x+3$. Using 1 rather than 31, I was also prompted to consider the area under the curve for $x^5+y^5=1$ between 0 and 1. See Figure 4.

Figure 4

In this graph, the vertical tangent occurs at $x=1$ and this got me thinking about the area under the curve between 0 and 1. In other words, what is the value of:$${\int }_0^1\sqrt[5]{1-x^5}dx$$Using SageMathCell, the integral evaluates to:$$0.2\cdot \beta (0.2,1.2)$$Now I vaguely remember coming across a so-called $\beta$ function that is in some way, I think, related to the $\mathrm{\Gamma }$ function. Further investigation revealed that they are indeed related. In fact:$$0.2\cdot \beta (0.2,1.2)=0.2\cdot \frac{\mathrm{\Gamma }(0.2)\cdot \mathrm{\Gamma }(1.2)}{\mathrm{\Gamma }(0.2+1.2)}\approx 0.950150138988437$$This result looks about right but what is going on with this $\beta$ function? Well, the beta function is actually an integral between 0 and 1 involving two different values, let's say $a$ and $b$, so that we have:$$\beta (a,b)={\int }_0^1{t}^{a-1}\cdot (1-t{)}^{b-1}dt$$If we can get our integral into that form then we can integrate it. Firstly, we make a substitution. Let $t=x^5$ and so $dt/dx=5x^4$ and $dx=0.2x^{-4}$. We can write $x^{-4}=t^{-0.8}$ and the integral now transforms into:$$\begin{array}{ll}{\displaystyle {\int }_0^1\sqrt[5]{1-x^5}dx}& =\text{0}.2{\displaystyle {\int }_0^1{t}^{-0.8}(1-t{)}^{0.2}dt}\\ & =0.2{\displaystyle {\int }_0^1{t}^{0.2-1}(1-t{)}^{1.2-1}dt}\\ & =0.2\beta (0.2,1.2)\end{array}$$So this was quite an interesting post that led me in some unexpected directions. I also was interested in the area between the curves $x^2+y^2=1$ and $x^5+y^5=1$ in the range from 0 to 1. See Figure 5 where eq2 is $x^2+y^2=1$ and the curve $x^5+y^5=1$ is a tangent to the circle at $x=0$ and $x=1$.

Figure 5

Clearly, the area between the two curves from $x=0\text{ to }1$ is given by $\beta (0.2,1.2)-{\displaystyle \frac{\pi }{4}}$. What I've learned from all this is that the $\beta$ function is a very useful tool to have in one's integration armoury.