Crazy integral with nested radicals and inverse sines

Question

Recently a friend who is writing a book on integrals added this problem to his book:

$$\int_{0}^{1}\arcsin{\sqrt{1-\sqrt{x}}}\ dx=\frac{3\pi}{16}$$

After a while, when trying to generalize, I was able to find the following integrals:

$$\int_{0}^{1}\arcsin{\sqrt{1-\sqrt{1-\sqrt{x}}}}\ dx=\frac{61\pi}{256}$$ $$\int_{0}^{1}\arcsin{\sqrt{1-\sqrt{1-\sqrt{1-\sqrt{x}}}}}\ dx=\frac{12707\pi}{65536}$$ For this last integral, simply apply a u-substitution with the term inside the arcsin: $$\int_{0}^{1}16u(1-u^2)(1-(1-u^2)^2)(1-(1-(1-u^2)^2)^2)\arcsin{u}\ du$$ Applying integration by parts and some algebra: $$\int_{0}^{1}\frac{(1-(1-(1-u^2)^2)^2)^2}{\sqrt{1-u^2}}\ du$$ With $\sin{t} = u$, the integral becomes: $$\int_{0}^{\frac{\pi}{2}}(1-(1-\cos^4{t})^2)^2\ dt$$ Which can be easily calculated with some formulas for the beta function. However, when trying to solve the problem for n radicals, I was not able to solve this last step in a more general way, furthermore, the integrals with more radicals quickly become too large to calculate manually. Is it possible to calculate integrals for n radicals with some more simplified expression?

Answer 1

Define $g_n(x)=\sqrt{1-\sqrt{1-\cdots\sqrt{1-\sqrt{x}}}}$, where there are $n$ radicals present. For instance $g_2(x)=\sqrt{1-\sqrt{x}}$ and $g_1(x) = \sqrt{x}$. Notice that for $n\ge 2$, $g_n(x)=g_{n-1}(1-\sqrt{x})$. Also define $$ I_{n,m} = \int_0^1 x^m \sin^{-1}(g_n(x))dx \overset{1-\sqrt{x}=t}{=} 2\int_0^1 (1-t)^{2m+1}\sin^{-1}(g_{n-1}(t))dt. $$ A recurrence relation is formed and we obtain $$ I_{n,m} = 2\sum_{r=0}^{2m+1}\binom{2m+1}{r}(-1)^rI_{n-1,r},$$ where we keep reducing until we reach terms of the form $I_{1,l}$ for $l\in \mathbb{N}$. This integral can be computed exactly, $$ I_{1,m} = \int_0^1 x^m \sin^{-1}(\sqrt{x})dx = 2\int_0^1 t^{2m+1}\sin^{-1}(t)dt = \frac{\pi}{2(m+1)}\left(1-\frac{\Gamma(m+3/2)}{\sqrt{\pi} (m+1)!} \right)$$ This also shows that the integral $I_{n,0}$ is always of the form $\frac{p}{q}\pi$ where $p$ and $q$ are rational numbers.

Addendum : An explanation of how I got the final integral. Using integration by parts and this result, $$\begin{align*}\int_0^1 t^{2m+1}\sin^{-1}(t)dt &= \left.\sin^{-1}(t) \frac{t^{2m+2}}{2m+2}\right|_{0}^1 -\frac{1}{2(m+1)}\int_0^1 \frac{t^{2m+2}}{\sqrt{1-t^2}}dt\\ &= \frac{\pi}{4(m+1)} - \frac{1}{2(m+1)} \cdot \sqrt{\pi}\frac{\Gamma(m+3/2)}{2\Gamma(m+2)} \\ &= \frac{\pi}{4(m+1)}\left(1-\frac{\Gamma(m+3/2)}{\sqrt{\pi} (m+1)!} \right).\end{align*}$$ For additional clarity on how the integrals are calculated, $$I_{1,0} = \frac{\pi}{2}\left[1-\frac{\Gamma\left(\frac{3}{2}\right)}{\sqrt{\pi}\cdot 1!}\right] = \frac{\pi}{2}\left[1-\frac{1}{2}\right] = \frac{\pi}{4}\\ I_{1,1} = \frac{\pi}{4}\left[1-\frac{\Gamma\left(\frac{5}{2}\right)}{\sqrt{\pi} \cdot 2!}\right] = \frac{\pi}{2}\left[1-\frac{3}{8}\right] = \frac{5\pi}{32} \\ I_{1,2}= \frac{\pi}{6}\left[1-\frac{\Gamma\left(\frac{7}{2}\right)}{\sqrt{\pi} \cdot 3!}\right] = \frac{\pi}{2}\left[1-\frac{5}{16}\right] = \frac{11\pi}{96} \\ I_{1,3}= \frac{\pi}{8}\left[1-\frac{\Gamma\left(\frac{9}{2}\right)}{\sqrt{\pi} \cdot 4!}\right] = \frac{\pi}{2}\left[1-\frac{35}{128}\right] = \frac{93\pi}{1024} $$ From these four integrals, $$I_{2,0} = 2I_{1,0}-2I_{1,1} = 2\cdot \frac{\pi}{4}-2\cdot \frac{5\pi}{32} = \frac{3}{16}\pi.$$ and $$I_{3,0} = 2I_{2,0}-2I_{2,1} = \frac{3\pi}{8}-4\left[I_{1,0}-3I_{1,1}+3I_{1,2}-I_{1,3}\right] \\ = \frac{3\pi}{8} - \pi +\frac{\pi}{2}+\frac{93}{256}\pi = \frac{61\pi}{256}$$ You have a lot of patience if you have computed $I_{4,0}$ by hand.

Addendum 2: Since you have asked for $I_{n,0}$, using the recurrence relation, $I_{n,0} = 2(I_{n-1,0}-I_{n-1,1}) = 2(2(I_{n-2,0}-I_{n-2,1})-2(I_{n-2,0}-3I_{n-2,1}+3I_{n-2,2}-I_{n-2,3}))$, thus, $$ \boxed{I_{n,0} = 4\big[ 2I_{n-2,1}-3I_{n-2,2}+I_{n-2,3}\big]}$$

Answer 2

Typical case of a textbook deception: By inspection, the radical its a pure sine-form by

$$\sqrt{1-\sqrt x } = \sqrt{1-\cos^2\phi}=\sin \phi , \quad 0\le \phi \le \frac{\pi}{2}$$

$$dx \sin ^{-1}\left(\sqrt{1-\sqrt{x}}\right)\text{/.}\, \left\{x\to \cos ^4(\phi ),dx\to \frac{\partial}{\partial \phi } \cos ^4(\phi ) \ d\phi\right\}\ = \ - \left( \sin (2 \phi )+ \frac{1}{2} \sin (4 \phi )\right) \phi \ d\phi$$

Can be done by partial integration of the trigs

$$-\int\left(\sin (2 \phi )+\frac{1}{2} \sin (4 \phi )\right) \, \phi \ d\phi =\frac{1}{32}\ (-8 \sin (2 \phi )-\sin (4 \phi )+16 \phi \cos (2 \phi )+4 \phi \cos (4 \phi ))$$