# Solve $\int_0^2 \sqrt{x+\sqrt{x+\sqrt{x+\dotsb}}}\,dx$

I haven't seen this question, but if someone has, it would be very appreciated if you could send a link!

I've been very interested in the MIT Integration Bee, and one question that stood out to me was:

$$\int_0^2 \sqrt{x+\sqrt{x+\sqrt{x+\dotsb}}}\,dx$$

I tried rewriting a couple of ways to simplify it, but nothing seemed to help. According to the official MIT Integration Bee website, the answer is $$\frac{19}{6}$$

Thanks!

• Can you identify the infinite nesting radical with a closed form? – Sangchul Lee Sep 1 '17 at 2:46
• Hint: If $y=\sqrt{x+y}$, then what is $y$ in terms of $x$? – Shalop Sep 1 '17 at 2:49
• @Shalop Convergence still needs to be justified. – MathematicsStudent1122 Sep 1 '17 at 2:50
• @MathematicsStudent1122 Hence a comment, not an answer. – Shalop Sep 1 '17 at 2:53

To formally justify this integration, we can do the following. Put $$f_1(x) = \sqrt x$$ ,and for each $$n\ge 1$$, put $$f_{n+1}(x) = \sqrt{x + f_{n}(x)}$$. We want to show that $$f(x) = \lim_{n\to\infty}f_n(x)$$ exists and is integrable on $$[0,2]$$. We will show that $$f(x)$$ is continuous on $$(0,2]$$, and has a discontinuity at $$x = 0$$, where $$f(0) = 0$$. This is sufficient because a function $$f\colon[0,2]\to\mathbf R$$ is Riemann integrable if and only if it is bounded and continuous almost everywhere. Finally, by "correcting" the value of $$f$$ at $$x=0$$ to produce a continuous function $$\tilde f$$ on $$[0,2]$$, we can evaluate the integral $$\int_0^2f(x)\,\mathrm dx$$ by applying the fundamental theorem of calculus to integrate $$\int_0^2\tilde f(x)\,\mathrm dx$$.

For each $$x\in (0,2]$$, put $$x_n = f_n(x)$$. We claim that $$x_n\to \frac{1}{2} + \frac{1}{2}\sqrt{4x+1}$$. First we will justify the convergence of the sequence $$(x_n)$$ using the monotone convergence theorem and then use algebraic limit laws to deduce its limit rigorously.

Consider the sequence $$b_{n+1} = \sqrt{2 + b_n}$$, where $$b_1 = \sqrt 2$$. Clearly $$b_1 \leqslant 2$$. Suppose that $$b_n\leqslant 2$$; then $$\sqrt{2 + b_n} \leqslant \sqrt{2 + 2} = 2$$. By induction, $$b_n \leqslant 2$$ for every $$n$$. Thus the sequence $$(b_n)$$ is bounded above by $$2$$. The sequence $$(b_n)$$ is clearly monotone increasing. Hence $$(b_n)$$ converges.

Since $$(x_n)$$ is monotone and bounded above by $$\lim_{n\to\infty}b_n$$, it converges, so set $$L = \lim_{n\to\infty}x_n$$. Since $$(x_n)$$ converges to $$L$$, every subsequence converges to $$L$$, so by continuity of the map $$y\mapsto y^2$$, we have $$L^2 = \big(\lim_{n\to\infty}x_{n+1}\big)^2 = \lim_{n\to\infty}x_{n+1}^2 = \lim_{n\to\infty}x + x_n = x + L.$$ Therefore $$L$$ is a root of the polynomial $$y\mapsto y^2 - y - x$$, and we can conclude that $$L = \frac{1}{2} + \frac{1}{2}\sqrt{4x+1}.$$ (Note that $$L$$ could not be the negative root since $$x_n \geqslant 0$$ for every $$n\geqslant 1$$ and every $$x\in(0,2]$$.)

If $$x = 0$$, then $$x_n = 0$$ for each $$n$$, so all told, $$f(x) = \begin{cases} \frac{1}{2} + \frac{1}{2}\sqrt{4x+1}, &\text{if x\in (0,2],}\\ 0, &\text{if x = 0.} \end{cases}$$

Thus we have proved that the integrand $$f(x)=\sqrt{x+\sqrt{x+\dotsb}}$$ is bounded and continuous almost everywhere on $$[0,2]$$, so it is Riemann integrable on $$[0,2]$$. The function $$\tilde f\colon [0,2]\to\mathbf R$$ defined by $$\tilde f(x) = \frac{1}{2} + \frac{1}{2}\sqrt{4x+1},\quad\text{for each x\in[0,2],}$$ is continuous on $$[0,2]$$ and agrees with $$f$$ almost everywhere, so $$\int_0^2 f(x)\,\mathrm dx = \int_0^2\tilde f(x)\,\mathrm dx$$. Since $$\tilde f(x)$$ is continuous on the entire closed interval $$[0,2]$$, its integral can be evaluated via the fundamental theorem of calculus. Since $$F(x) = \frac{x}{2} + \frac{(4x+1)^{3/2}}{12}$$ satisfies $$F'(x) = \tilde f(x)$$ for each $$x\in [0,2]$$, by the fundamental theorem of calculus, \begin{align*} \int_0^2\tilde f(x)\,\mathrm dx = F(2)-F(0) = \bigg[\frac{2}{2} + \frac{(4\cdot 2+1)^{3/2}}{12}\bigg] - \bigg[\frac{0}{2}+\frac{(4\cdot0+1)^{3/2}}{12}\bigg]=\frac{19}{6}, \end{align*} as desired.

• Just to elaborate, the sequence $b_n$ is monotone increasing because $u \leq \sqrt{u+2}$ whenever $|u| \leq 2$ – Shalop Sep 1 '17 at 17:32

Informally, if we set the nested radical equal to $y$ then an observation to make is that we have $$y = \sqrt{x+y}.$$ This leads us to $$y^2-y-x=0.$$ Looking at the positive solution to this, we have $$y = \frac{1}{2}(\sqrt{4x+1}+1).$$ So, we integrate the $y$ that we just found over the interval from 0 to 2 to achieve $\frac{19}{6}.$

Note that this is very informal. I suspect that justifying convergence in the middle of MIT's integration bee is not necessary.

• For the sake of justification, how would you do it? – John Lou Sep 1 '17 at 3:00
• @JohnLou I don't know for sure but my guess is that one would set up a recursive formula $x_{n+1}=\sqrt{x+x_n}$ and use monotone convergence. Perhaps later, I could try to come up with a full argument. I'm not entirely sure what the initial value would be in this case. – user328442 Sep 1 '17 at 3:03
• How can you justify that we have to take the positive quadratic (as opposed to $1-\sqrt{4x+1}$? – John Lou Sep 1 '17 at 3:06
• I assumed that to be the case since the interval of interest contains value of x from 0 to 2. If we consider these values then y ranges between 0 and -1, which doesn't seem appropriate given a nested square root. – user328442 Sep 1 '17 at 3:08