# Unsure of my work evaluating $\int \frac{dx}{\sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}}$

This Question is an Extension of this Previously Asked Question: Nested root integral $\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x}}}}$

I was looking into answering the question of whether it was possible to integrate the fully nested root integral of the variety described in the previous problem: $$\int \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}$$ So I started by defining the nested root in another way $$u=\sqrt{x+u} \therefore \\ u^2-u=x \\ (2u-1)du = dx$$ Using the results of the substitution I have set up $$\int \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \int \frac{2u-1}{u} du \\ = \int \left(2-\frac{1}{u}\right)du = 2u-\ln(u)$$

I am unsure of my work, as I have never attempted to integrate any infinitely nested functions. Therefore I have no idea whether my method for u-substitution is valid. Am I just living under a rock or have other people seen this method used previously? For such a seemingly intimidating problem it was surely quite easy.

I had difficulty checking my work using wolfram alpha, but I managed to confirm that this works for the definite integral limits from $$x = 1$$ to $$x = 2$$ and from $$x = 1$$ to $$x = 3$$. Maybe I am just flat out wrong and got lucky on these two calculations?

• Your method seems very clear, but one thing is not very clear to me. I can not find any resource which confirm or denies if substitution rule can be used for such "recursive" substitutions, but you seemed to have unwound it to a simple substitution. Intuitively this should be correct. Very interesting question indeed. – TheCoolDrop Jul 25 at 7:24

Your work is perfectly fine if your lower and upper integral limits satisfy $$0 < a \leq b$$. In that case your answer $$2u - \ln(u)$$ even has a nice closed form purely in terms of $$x$$:

$$\int_a^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \Big[\big(1 + \sqrt{4x + 1}\big) - \ln\Big(\frac{1}{2}\big(1 + \sqrt{4x + 1}\big)\Big)\Big]_a^b$$

This formula also continues to work for a lower limit of $$a = 0$$ if you interpret either the integral and/or the nested radical $$\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$$ in the denominator properly enough.

Since you used $$u$$-substitution, your method should work as long as the conditions for an integration by substitution are met. Say you are integrating over some interval $$[a, b]$$. You have to verify:

• Does the function $$u(x) = \sqrt{x + u(x)}$$ that you defined implicitly actually make sense over $$[a, b]$$? In other words, is there really a function $$u : [a, b] \to \Bbb R$$ that satisfies that recursion?

• Is the function $$u(x)$$ actually differentiable over $$[a, b]$$?

$$\underline{\textit{There is some good news for these questions:}}$$

As long as $$x > 0$$, there is a well-defined expression for $$u$$ in terms of $$x$$ when $$u = \sqrt{x + u}$$. To realize this, we need translate the intuitive expression $$\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$$ into the precise language of calculus. Only then can we bring the full power of calculus to bear on this problem. So formally what is going on with a nested radical like $$\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$$ is this:

Let $$u_{x,1} = \sqrt{x}$$ and define recursively the sequence $$u_{x,n + 1} = \sqrt{x + u_{x,n}}$$ ($$n \in \Bbb Z_+$$). If $$u_x = \lim\limits_{n \to \infty}u_{x,n}$$ exists, then we may define our sought-after function $$u$$ at $$x$$ to be $$u(x) = u_x$$. In essence, the limit $$\lim\limits_{n \to \infty}u_{x,n}$$ is mathematically what we define the expression $$\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$$ to be. And we can easily check that $$u_x = \sqrt{x + u_x}$$ by taking the limit as $$n \to \infty$$ at both sides of the equation $$u_{x,n + 1} = \sqrt{x + u_{x,n}}$$.

Now the good news is that, as long as $$x > 0$$, you can show that the sequence $$u_{x,n}$$ is bounded and monotonically increasing so that it does converge to a definite limit, namely

$$u_x = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$$

Hence, our function $$u(x) = u_x$$ is well-defined for $$x > 0$$. Also, note that the formula above should not surprise you. You can easily see where it originated:

Informally, if you take your substitution equation $$u^2 - u = x$$ and wrote it as a quadratic equation $$u^2 - u - x = 0$$, you can solve it by thinking of $$x$$ as a constant. And indeed, one of the solutions that pops out is precisely $$u_+ = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$$. You can eliminate the other solution $$u_- = \frac{1}{2}\big(1 - \sqrt{4x + 1}\big)$$ since it is negative if $$x > 0$$ and by convention square roots are positive.

So as long as $$x > 0$$, you can safely take $$u(x) = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$$ as the $$u$$-substitution function which satisfies $$u = \sqrt{x + u}$$. In fact, as is apparent from the formula, $$u(x)$$ is even differentiable in this case.

$$\underline{\textit{But there are caveats:}}$$

$$1.\ \textbf{Note that for x < 0, the limit does not make sense:}$$ as the very first sequence element $$u_{x,1} = \sqrt{x}$$ is not real. So, from this very analysis, you can immediately conclude that you should not be integrating over negative values in your integral.

$$2.\ \textbf{Next, at x = 0, things almost work out but break down anyway:}$$ Note, we only managed to eliminate $$\frac{1}{2}\big(1 - \sqrt{4x + 1}\big)$$ as a candidate for the limit above because it was negative if $$x > 0$$. Well, if $$x = 0$$, then $$u_- = \frac{1}{2}\big(1 - \sqrt{4x + 1}\big) = 0$$ and you can no longer eliminate it that easily. So, we must go back to our definition of $$\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$$ in terms of sequences to arbitrate between $$u_+$$ and $$u_-$$. Applying that definition when $$x = 0$$, we see that $$u_- = 0$$ is the candidate that is chosen this time not $$u_+ = 1$$. This is because in this case, all the sequence elements $$u_{x,n}$$ are zero: $$u_{x=0,1} = \sqrt{x} = \sqrt{0} = 0,\quad u_{x=0,2} = \sqrt{x + u_{x=0,1}} = \sqrt{0 + 0} = 0,\quad \ldots \text{ etc}$$

Hence, $$\lim\limits_{n \to \infty}u_{x=0,n} = 0$$ and $$u(0) = 0$$. However, approaching $$0$$ from the right, we see that $$\lim\limits_{x \to 0+}u(x) = \frac{1}{2}\big(1 + \sqrt{4\cdot0 + 1}\big) = 1$$ And therefore even though $$u(x)$$ is defined at $$x = 0$$, it is sadly not continuous there, let alone differentiable. So the $$u$$-substitution Theorem no longer applies.

In any case, there is an even worse problem when $$x = 0$$. Note that the function you are trying to integrate $$f(x) = \sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$$ is undefined at $$x = 0$$ because as we saw, our definition of $$\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$$ in terms of sequences gives you a $$0$$ when $$x = 0$$. So there would be a $$0$$ in your denominator for $$f(x)$$ if that was allowed.

$$\underline{\textit{Okay, so we have concluded so far that:}}$$

As long as your integration interval $$[a, b]$$ satisfies $$0 < a \leq b$$, your work should go through and you can use $$u(x) = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$$ as the explicit formula for $$u$$ to express your final integral answer:

$$\int_a^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \Big[\big(1 + \sqrt{4x + 1}\big) - \ln\Big(\frac{1}{2}\big(1 + \sqrt{4x + 1}\big)\Big)\Big]_a^b$$

$$\underline{\textit{Fixing the breakdown at x = 0:}}$$

If you really want $$x = 0$$ as one of the limits e.g. $$\int_0^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}}$$ for $$b > 0$$, you can do so in two ways, both of which lead to the same result:

• You can modify the definition of $$\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$$ thus: it defaults to the usual definition via sequences if $$x > 0$$ and to $$\frac{1}{2}(1 + \sqrt{4 \cdot 0 + 1}) = 1$$ if $$x = 0$$. Then you can safely use $$u(x) = \frac{1}{2}\big(1 + \sqrt{4x + 1}\big)$$ for all $$x \geq 0$$. And the answer you will get for your integral is exactly what you would expect by plugging in $$a = 0$$ in the closed form I gave above: $$\big(1 + \sqrt{4b + 1}\big) - \ln\Big(\frac{1}{2}\big(1 + \sqrt{4b + 1}\big)\Big) - 2$$

• On the other hand, you can instead take a limiting integral in the same spirit that we define $$\int_0^b \frac{1}{x^2}dx$$ to get around the singularity of $$\frac{1}{x^2}$$ at $$0$$. That is, you can define: \begin{align*} \int_0^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} &:= \lim_{a \to 0+}\int_a^b \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} \\ &= \lim_{a \to 0+}\big[2u(x) - \ln(u(x))\big]_a^b \end{align*} This leads to the same answer because ultimately $$\lim\limits_{a \to 0^+}u(x) = 1$$.

• Posts like this make me come back. Great answer, very informative. – TheCoolDrop Jul 25 at 8:50
• @TheCoolDrop Glad to be of help. – 0XLR Jul 25 at 9:34