A double integral with functions as bounds: $\int_0^1 dx \int_{\frac{\sqrt[3]{x}}2}^{\sqrt[3]{x}} \sqrt{1-y^4}dy$

Question

So I have the integral

$$\int_0^1 dx \int_{\frac{\sqrt[3]{x}}{2}}^{\sqrt[3]{x}} \sqrt{1-y^4}dy.$$

How on earth do I do this integral? I used WolframAlpha to have a look at what $\sqrt{1-y^4}$ integrates to, and it's absurd. I've tried quite a few different tricks and substitutions, as well as swapping the order which doesn't change anything, I feel like I'm missing something really obvious.

Thanks!

Edit: This was a practice question for a maths course I'm currently in. Turns out our lecturer chose the wrong bounds for what he had in mind - the solution for it is certainly outside the scope of the course!

Answer 1

We have

$$ \int_0^1 dx \int_{\frac{\sqrt[3]{x}}{2}}^{\sqrt[3]{x}}\sqrt{1-y^4}\:dy=\frac76-\frac{5\sqrt{15}}{16}+\frac{\sqrt{\pi}}8\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{7}{4}\right)}-\frac12 \: _2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};\frac1{16}\right). $$

Proof. By integrating by parts, one has $$ \begin{align} I:=\int_0^1 dx \int_{\frac{\sqrt[3]{x}}{2}}^{\sqrt[3]{x}} \sqrt{1-y^4}\:dy=\left[x\cdot \int_{\frac{\sqrt[3]{x}}{2}}^{\sqrt[3]{x}}\sqrt{1-y^4}\:dy\right]_0^1-\int_0^1x\cdot v'(x)\: dx \end{align} $$ with $$ v'(x)=\frac{d}{dx}\int_{\frac{\sqrt[3]{x}}{2}}^{\sqrt[3]{x}} \sqrt{1-y^4}\:dy=\frac{\sqrt{1-x^{4/3}}}{3x^{2/3}}-\frac{\sqrt{16-x^{4/3}}}{24x^{2/3}} $$ where we have used the Leibniz integral rule. Thus $$ \begin{align} I&=\int_{\large \frac12}^1\sqrt{1-y^4}\:dy\:-\frac13\int_0^1x^{1/3}\sqrt{1-x^{4/3}}\:dx+\frac1{24}\int_0^1x^{1/3}\sqrt{16-x^{4/3}}\:dx \\\\&=I_1-I_2+I_3. \end{align} $$ By the change of variable $u=x^{4/3}$ we clearly get $$ I_2=\frac13\int_0^1x^{1/3}\sqrt{1-x^{4/3}}\:dx=\frac14\int_0^1\sqrt{1-u}\:du=\frac16 $$ and $$ I_3=\frac1{24}\int_0^1x^{1/3}\sqrt{16-x^{4/3}}\:dx=\frac1{32}\int_0^1\sqrt{16-u}\:du=\frac{4}{3}-\frac{5\sqrt{15}}{16}. $$ Then we may write $$ \begin{align} I_1&=\int_{\large \frac12}^1\sqrt{1-y^4}\:dy=\int_0^1\sqrt{1-y^4}\:dy-\int_0^{\large \frac12}\sqrt{1-y^4}\:dy \\\\&=\frac{\sqrt{\pi}}8\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{7}{4}\right)}-\int_0^{\large \frac12}\sqrt{1-y^4}\:dy \\\\&=\frac{\sqrt{\pi}}8\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{7}{4}\right)}+\int_0^{\large \frac12}\sum _{n=0}^{\infty } \frac{\Gamma\left(n-\frac12\right)}{2\cdot\Gamma\left(\frac12\right)\cdot n!}\cdot\left(-y^4\right)^n\:dy \\\\&=\frac{\sqrt{\pi}}8\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{7}{4}\right)}-\sum _{n=0}^{\infty } (-1)^n\frac{\Gamma\left(n-\frac12\right)}{2\cdot\Gamma\left(\frac12\right)\cdot n!}\cdot\frac1{2^{4n+1}(4n+1)} \\\\&=\frac{\sqrt{\pi}}8\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{7}{4}\right)}-\frac12 \: _2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};\frac1{16}\right) \end{align} $$ where we have used the Euler beta function and the classic power series expansion for $_2F_1\left(a,b;c;x\right)$. Inserting $I_1, I_2, I_3$ into $I$ gives the announced result.