Computing $\int \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\cdots}}}} \,\mathrm{d}x$

Question

I've just read a post that has been put on hold. The question was about computing $$ \int_{ }^{ }\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x \cdots}}}}\, \mathrm{d}x $$ My attempt was to find explicitely the integrand hence I introduced the sequence $$ a_1=1\ \text{ and }\ a_{n+1}=\left(xa_n\right)^{1/(n+1)} $$ hence $$ a_2=\sqrt{xa_1}=\sqrt{x}, \ a_3=\sqrt[3]{xa_2}=\sqrt[3]{x\sqrt{x}} \ \dots $$ But the power is reversed, how can I ffind the corresponding $(a_n)_{n\in \mathbb{N}}$ ?

Answer 1

$\sqrt{x\sqrt[3]{x\sqrt[4]{x{...\sqrt[n]{x}}}}} = \sqrt{x} \cdot \sqrt{\sqrt[3]{x}} \cdot \sqrt{\sqrt[3]{\sqrt[4]{x}}} \cdot... = \displaystyle{x^{\sum_{k=2}^n\frac{1}{k!}}}$.

Let $r = \displaystyle{{\sum_{k=2}^n\frac{1}{k!}}}$. So $\displaystyle{\int{x^rdx} = \frac{x^{r+1}}{r+1} + \mathcal{C}}$.