Compute the indefinite integral

$$\displaystyle \int\sqrt{x\sqrt{x\sqrt{x\cdots\sqrt{x}}}}~~dx=\displaystyle \int\sqrt{x\underset{n~\text{times}}{\underbrace{\sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}}}}~~dx$$ where $x$ is repeated $n$ times.

My Attempt (I hope I have succeeded in the solution)

$1)$ We put $$\displaystyle y_{n}=\sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}$$

so we have

$$\displaystyle y_{n}=x^{\frac{2^{n}-1}{2^{n}}}=x^{1-\frac{1}{2^{n}}}$$

$$\displaystyle \int y_{n}dx=\frac{x^{2-\frac{1}{2^{n}}}}{2-\frac{1}{2^{n}}}+c$$

$2)$ Important note $$\displaystyle \lim_{n \to \infty }\displaystyle\displaystyle\int\sqrt{x\underset{n}{\underbrace{\sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}}}}~dx=\frac{x^{2}}{2}+c $$

You have all my respect and appreciation. Thank you!

  • $\begingroup$ The last equation in part 1 is wrong. The coefficient in front of $x$ is not $1/2$, unless $y_n=x$. Try for $y_0$ or for $y_1$. $\endgroup$
    – Andrei
    Aug 29, 2022 at 14:30
  • $\begingroup$ This certainly works. You could also separate the square roots and then sum $1/2 + 1/4 + 1/8 + ... + 1/2^n = 1-1/2^n$ and get the same answer. $\endgroup$ Aug 29, 2022 at 14:32
  • $\begingroup$ $y_{n}=x$ in order to $n\longrightarrow \infty$ $\endgroup$ Aug 29, 2022 at 14:34

1 Answer 1


Note that:$$\sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}=x^{\frac{1}{2}} x^{\frac{1}4} x^{\frac{1}8}\cdots x^{\frac{1}{2^n}}=x^{\frac{1}2+\frac{1}4+\cdots+\frac{1}{2^n}}=x^{1-\frac{1}{2^n}}$$

$$\int \sqrt{x\sqrt{x\sqrt{x...\sqrt{x}}}}~~ dx=\frac{1}{2-\frac{1}{2^n}}x^{2-\frac{1}{2^n}}+C$$


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