# Impossible Integration Problem?

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:

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

EDIT:

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.

• 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). Commented Sep 28, 2021 at 23:27
• 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? Commented Sep 28, 2021 at 23:27

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.
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.