Computing the integral $I=\int_{0}^{1}\sqrt{1-x^{\pi}}dx$

Question

Let $I=\int_{0}^{1}\sqrt{1-x^{\pi}}dx$ If we use $u=x^{\pi}$ then we get $dx=\frac{du}{\pi^{\frac{\pi-1}{\pi}}}$ are there any better identites or substitution that can make integral easier to compute?

Answer 1

[Note] : You can use the beta function :

• Beta Function : $\displaystyle \mathfrak{B}(x,y):=\int_{0}^{1}(1-\mu)^{x-1}\mu^{y-1}\;\text{d}\mu=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

Let $\mu:=x^{\pi}$. Then : \begin{align*} x^{\pi}=\mu&\implies\text{d}\mu=\pi x^{\pi -1}\\ \\ &\iff\text{d}x=\frac{\text{d}\mu}{\pi\cdot\mu^{\displaystyle\frac{\pi-1}{\pi}}} =\frac{1}{\pi}\mu^{\displaystyle\frac{1-\pi}{\pi}}\text{d}\mu \end{align*} Therefore : \begin{align*} I&=\frac{1}{\pi}\int_{0}^{1}\sqrt{1-\mu}\cdot\mu^{\displaystyle\frac{1-\pi}{\pi}}\text{d}\mu\\ \\ &=\frac{1}{\pi}\int_{0}^{1}(1-\mu)^{\frac{3}{2}-1}\mu^{\frac{1}{\pi}-1}\;\text{d}\mu\\ \\ &=\frac{1}{\pi}\mathfrak{B}\left(\frac{3}{2},\frac{1}{\pi}\right)\\ \\ &=\frac{1}{\pi}\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{\pi}\right)}{\Gamma\left(\frac{3}{2}+\frac{1}{\pi}\right)}\\ \\ &\approx 0.846941\cdots \end{align*}

Answer 2

Just for the fun of it

As said in comments and answers, the is no closed form expression for the antiderivative. However, we could have an infinite series representation of it since $$\sqrt{1-x^{\pi}}=\sqrt{1-x^{3}}\Bigg[1+x^3\sum_{n=1}^\infty \frac { 1} {2^n\, n! }\left(\frac{\log (x)}{x^3-1}\right)^n P_n(x)\,(\pi-3)^n\Bigg] $$ where the first $P_n(x)$ are $$\left( \begin{array}{cc} n & P_n(x) \\ 1 & 1 \\ 2 & x^3-2 \\ 3 & x^6-2 x^3+4 \\ 4 & x^9-4 x^6-4 x^3-8 \\ 5 & x^{12}-4 x^9+36 x^6+56 x^3+16 \\ 6 & x^{15}-6 x^{12}-76 x^9-496 x^6-336 x^3-32 \\ 7 & x^{18}-6 x^{15}+288 x^{12}+3088 x^9+5328 x^6+1632 x^3+64 \end{array} \right)$$

For any $n$, each term shows an explicit antiderivative in terms of hypergeometric function. For example, for $n=1$ $$\int \frac{ x^3 \log (x)}{ \sqrt{1-x^3}}\,dx=$$ $$-\frac{x^4}{16} \left(\, _3F_2\left(\frac{1}{2},\frac{4}{3},\frac{4}{3};\frac{7}{3},\frac{7}{3};x^3\right)- 4 \, _2F_1\left(\frac{1}{2},\frac{4}{3};\frac{7}{3};x^3\right) \log (x)\right)$$

Using only the elements given in this answer, computing the integral from $0$ to $t$ gives the following results

$$\left( \begin{array}{cccc} t & \text{approximation} & \text{solution} & \text{error} \\ 0.1 & 0.099991285257983 & 0.099991285257855 & 1.29 \times 10^{-13}\\ 0.2 & 0.199846061391490 & 0.199846061391227 & 2.63 \times 10^{-13}\\ 0.3 & 0.299172693249054 & 0.299172693248698 & 3.56 \times 10^{-13}\\ 0.4 & 0.397263323235192 & 0.397263323234765 & 4.26 \times 10^{-13}\\ 0.5 & 0.493045151618189 & 0.493045151617706 & 4.83 \times 10^{-13}\\ 0.6 & 0.584997454713703 & 0.584997454713173 & 5.30 \times 10^{-13}\\ 0.7 & 0.670989777542907 & 0.670989777542338 & 5.70 \times 10^{-13}\\ 0.8 & 0.747926304221105 & 0.747926304220506 & 6.00 \times 10^{-13}\\ 0.9 & 0.810765563739254 & 0.810765563738633 & 6.20 \times 10^{-13}\\ 1.0 & 0.846941809210854 & 0.846941809208595 & 2.26 \times 10^{-12} \end{array} \right)$$

Otherwise and much simpler (but more difficult to compute) $$I=\int\sqrt{1-x^{\pi}}\,dx=\frac{x \left(\pi \, _2F_1\left(\frac{1}{2},\frac{1}{\pi };1+\frac{1}{\pi };x^{\pi }\right)+2 \sqrt{1-x^{\pi }}\right)}{2+\pi }$$