How to find $\lim_{n \to \infty} \int_0^1 \cdots \int_0^1 \sqrt{x_1+\sqrt{x_2+\sqrt{\dots+\sqrt{x_n}}}}dx_1 dx_2\dots dx_n$

Solution 1:

Let us define: \begin{equation} I_3(a) := \int\limits_{[0,1]^3} \sqrt{x+\sqrt{y+\sqrt{z+a}}} dx dy dz \end{equation} Then by using elementary integration we have the following result: \begin{eqnarray} I_3(a) &=& \frac{32}{31} \left( \sum\limits_{k=0}^3 |[\begin{array}{r} 3 \\ k \end{array}]|(-1)^{3-k} \frac{(u^+)^{\frac{9}{2}+k} - (u^-)^{\frac{9}{2}+k}}{\frac{9}{2}+k} \right)+\\ &-& \frac{32}{15}\left( \sum\limits_{k=0}^3 \binom{3}{k}(-1)^{3-k} \frac{(u_1^+)^{\frac{7}{2}+k} - (u_1^-)^{\frac{7}{2}+k}}{\frac{7}{2}+k} \right)+\\ &-&\frac{16}{21}\left( \frac{(u^+_2)^{\frac{15}{4}} - (u^-_2)^{\frac{15}{4}}}{\frac{15}{4}}-\frac{(u^+_2)^{\frac{11}{4}} - (u^-_2)^{\frac{11}{4}}}{\frac{11}{4}}-\frac{(u^+_3)^{\frac{15}{4}} - (u^-_3)^{\frac{15}{4}}}{\frac{15}{4}} \right) \end{eqnarray} where \begin{eqnarray} (u^+, u^-) &:=& (\sqrt{1+\sqrt{1+a}}+1,\sqrt{1+\sqrt{0+a}}+1) \\ (u_1^+,u_1^-) &:=& (\sqrt{0+\sqrt{1+a}}+1,\sqrt{0+\sqrt{0+a}}+1)\\ (u_2^+,u_2^-)&:=&(\sqrt{1+a}+1,\sqrt{0+a}+1)\\ (u_3^+,u_3^-)&:=&(\sqrt{1+a}+0,\sqrt{0+a}+0) \end{eqnarray} Now, clearly the next integral we need to compute is $I_4(a) := \int\limits_0^1 I_3(\sqrt{\xi+a}) d\xi$. All the integrals are can be expressed through elementary functions by substituting for the respective $u_j^{\pm}$ for $j=0,\cdots,3$ and integrating power functions. Therefore with some effort it is easy to compute higher elements of this sequence. Yet it is unclear for me at this stage if I might be able to find a neat expression for arbitrary elements.