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 $$ \frac{19}{6} $$


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

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    $\begingroup$ Just to elaborate, the sequence $b_n$ is monotone increasing because $u \leq \sqrt{u+2}$ whenever $|u| \leq 2$ $\endgroup$ – 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.

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    $\begingroup$ For the sake of justification, how would you do it? $\endgroup$ – John Lou Sep 1 '17 at 3:00
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    $\begingroup$ @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. $\endgroup$ – user328442 Sep 1 '17 at 3:03
  • $\begingroup$ How can you justify that we have to take the positive quadratic (as opposed to $1-\sqrt{4x+1}$? $\endgroup$ – John Lou Sep 1 '17 at 3:06
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    $\begingroup$ 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. $\endgroup$ – user328442 Sep 1 '17 at 3:08

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