A curious integral through an algebraic equation

Let us assume that $y > 0$ .

The substitutions $u^2 = \tfrac{1+y}{x} - y$ and $v = u - 1$ gives $A \; = \; \int_0^1 \sqrt{-1 + \sqrt{\frac{1+y}{x} - y}}\,dx \; = \; 2(1+y)\int_1^\infty \frac{u\sqrt{u-1}}{(u^2+y)^2}\,du \; = \; 2(1+y)\int_0^\infty \frac{(v+1)\sqrt{v}}{\big((v+1)^2+y\big)^2}\,dv$ Defining $\sqrt{v}$ on the cut plane $|v| > 0$ , $0 < \mathrm{Arg}\,v < 2\pi$ , and integrating around a suitable keyhole contour gives $A \;= \; 2\pi i(1 + y)\Big(\mathrm{Res}_{v=-1+i\sqrt{y}}\frac{(v+1)\sqrt{v}}{\big((v+1)^2 + y\big)^2} + \mathrm{Res}_{v=-1-i\sqrt{y}}\frac{(v+1)\sqrt{v}}{\big((v+1)^2 + y\big)^2} \Big)$ Now the function $v \; \mapsto \; f_y(v) \; = \; \frac{(v+1)\sqrt{v}}{\big((v+1)^2 + y\big)^2}$ has a double pole at each of $v = -1 \pm i\sqrt{y}$ , and hence $\mathrm{Res}_{y = -1+i\sqrt{y}}\frac{(v+1)\sqrt{v}}{\big((v+1)^2 + y\big)^2} \; = \; \frac{d}{dv} \frac{(v+1)\sqrt{v}}{(v + 1 + i\sqrt{y})^2} \Big|_{v=-1+i\sqrt{y}} \; = \; -\frac{\sqrt{-1+i\sqrt{y}}(\sqrt{y}-i)}{8\sqrt{y}(1+y)}$ and, similarly, $\mathrm{Res}_{y = -1-i\sqrt{y}}\frac{(v+1)\sqrt{v}}{\big((v+1)^2 + y\big)^2} \; = \; -\frac{\sqrt{-1-i\sqrt{y}}(\sqrt{y}+i)}{8\sqrt{y}(1+y)}$ With this form of the cut plane, we have $\sqrt{-i+i\sqrt{y}} = a + ib$ and $\sqrt{-1-i\sqrt{y}} = -a + ib$ for some $a,b > 0$ , and we deduce that $A \; = \; \frac{\pi}{2\sqrt{y}}(b\sqrt{y}-a)$ We now calculate that $a \; = \; \sqrt{\frac{\sqrt{1+y}-1}{2}} \hspace{2cm} b \; = \; \sqrt{\frac{y}{2\big(\sqrt{1+y}-1\big)}}$ obtaining, finally, $A \; = \; \pi\sqrt{\frac{(1+y)\big(\sqrt{1+y}-1\big)}{8y}}$ Putting $t = \sqrt{1+y}$ , we need to solve the equation $\begin{aligned} (1+y)\big(\sqrt{1+y}-1\big) & = \; 8y \\ t^2(t-1) & = \; 8(t^2-1) \\ (t-1)(t^2 - 8t - 8) & = \; 0 \end{aligned}$ Thus we must have $t=1\,,\, 4 \pm 2\sqrt{6}$ . Since $y > 0$ we must have $t > 1$ and hence $t = 4+2\sqrt{6}$ . Thus $y = 39 + 16\sqrt{6}$ , and hence $y^2 - 78y = \boxed{15}$ .

Note that although $t=1$ is a solution of the final cubic equation, $y=0$ is not a solution to this problem. The formula for $A$ is not valid for $y=0$ , and indeed the limit of $A$ as $y \to 0$ is $\tfrac14\pi$ , which is the correct value of the integral when $y=0$ , as an $x = \sin^4\theta$ substitution readily shows). By its definition, $t$ must be positive, and so the $4 - 2\sqrt{6}$ value must be discarded from this analysis.

That said, there is a second solution for $y$ , where $-1 < y < 0$ . If we put $y = -p^2$ where $0 < p < 1$ , then the poles for $f_y(v)$ are $-(1+p)$ and $-(1-p)$ , both real and negative. The same contour integration can be attempted, and in this case we obtain $A \; = \; \pi\frac{\sqrt{1-p^2}}{4p}\big(\sqrt{1+p} - \sqrt{1-p}\big)$ and we obtain $A=\pi$ when $p^4 + 78p^2 - 15 = 0$ , or $y^2 - 78y - 15 = 0$ , and hence we obtain a solution for $y$ with $-1 < y < 0$ , namely $y = 39 - 16\sqrt{6}$ . This second solution still gives $y^2 - 78y = 15$ , of course.