Impossible Integration Problem?

Question

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.

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.

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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.

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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 :-)