Let $f(x, y) = \sqrt{x + y\sqrt{x + y\sqrt{...}}}$

Evaluate $\int_0^2 \int_0^2 f(x, y) \,dy \,dx$

I gave this problem to a few of my friends and my Calculus teacher and no one could solve it, nor could any online math tool/solver, although the antiderivative must surely exist.

Everyone I posed the problem to could get roughly up to here:

First, we say $Z = f(x, y)$, and with some basic algebraic manipulations, you can say:

$$ Z = \sqrt{x + y\sqrt{x + y\sqrt{...}}} \\ Z^2 = x + yZ \\ Z^2 - yZ - x = 0 $$

Now you have a quadratic. You can complete the square, or what's easier in my opinion is to simply substitute into the quadratic formula:

$$ \begin{aligned} Z &= \frac{-(-y) \pm \sqrt{(-y)^2 - 4(1)(-x)}}{(2)(1)} \\ &= \frac{y \pm \sqrt{y^2 + 4x}}{2} \end{aligned} $$

Since $Z$ must be positive, as the square root is only defined in the reals for positive inputs/outputs, the $\pm$ can only be $+$ (since $y^2 + 4x$ is strictly greater than $y$, so subtracting $y - \sqrt{y^2 +4x}$ would be negative.

Subsequently, you have a function $Z$ with a finite number of terms that is equivalent to the original $f(x, y)$. So, you might try to integrate it.

However, this function has no easy antiderivative, and no one I know nor any resource I've found has been able to solve it (without simply naively approximating the definite integral with a computer).

Is it possible to evaluate this definite integral with conventional methods. Furthermore, is it possible to evaluate the indefinite integral?

Just to be clear, the problem ultimately simplifies to the following:

$$ \int_0^2 \int_0^2 \frac{y + \sqrt{y^2 + 4x}}{2} \,dy \,dx $$


I forgot to mention that Wolfram Alpha does give a result for the indefinite integral, but showing the steps exceeds the computational time constraints of the free account tier. It says the solution is:

$$ \frac{1}{48}\big(24x^2\log\big(\sqrt{4x + y^2} + y\big) - 6x^2 + 2xy\big(5\sqrt{4x + y^2} + 6y\big) + y^3\big(\sqrt{4x + y^2} + 3y\big)\big) $$

This is a long expression and I don't suppose it's obvious how there is a logarithm in the antiderivative.

  • 3
    $\begingroup$ First you should prove that the limit of this nested square roots really exist. For the final integral it is probably easier to integrate first in $x$ (using Tonelli's theorem). $\endgroup$ Commented Sep 28, 2021 at 23:27
  • 5
    $\begingroup$ So it's $$\frac 12 \left (\int_0^2 \int_0^2 y~dy~dx + \int_0^2 \int_0^2 \sqrt{y^2+4x}~dy~dx \right ).$$ Aren't both summands relatively easy to compute? $\endgroup$ Commented Sep 28, 2021 at 23:27

1 Answer 1


We are essentially trying to evaluate the following integral, as you have observed: $$\int_0^2 \int_0^2 \frac{y + \sqrt{y^2 + 4x}}{2} \,dy \,dx.$$ I'll focus on finding the indefinite integral $$I=\int \sqrt{y^2+4x}~dy$$ as this seems to be your main area of difficulty, and then hopefully you can complete the rest of your problem.

You can use hyperbolic substitution here, you can also use some clever manipulations that I personally would enjoy doing but you can also use some straightforward trig substitutions which is somewhat more low-tech (which in no way means you shouldn't use it! Often the most basic tricks are the best).

Let $y=2\sqrt x \tan u$. This means that $dy=2\sqrt x \sec^2 u~du$. Hence, $$\begin{align}I&=\int \sqrt{(2\sqrt x\tan u)^2+4x}~\cdot2\sqrt x\sec^2u~du\\ &=\int\sqrt{4x(\tan^2u+1)}\cdot2\sqrt x\sec^2u~du\\ &=\int4x\sec^3u du=4x\int\sec^3u~du\end{align}$$ where I've used the identity $1+\tan^2u\equiv \sec^2u$ in the $3$rd line. The integral of $\sec^3u$ with respect to $u$ is well known and can be quickly found with some sweet manipulations. For example, see here: Wikipedia (Integral of Secant Cubed) . It follows that $$I=2x\sec u\tan u+2x\log\lvert \sec u+\tan u\rvert +C.$$ You can now use the identity quoted above to make this in terms of $y$, and the definite integral will be solved.

I hope that helps. If you have any questions (such as the motivation behind the substitution, what manipulation I'm referring to that I say I'd enjoy doing or anything else) please don't hesitate to ask :-)


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