Showing $\int_1^\infty\left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)dx=\frac4{15}\left(\sqrt{26\sqrt2-14}-2\right)$ A Putnam problem asked to show that some improper  integral is convergent, but I was curious to see if it can be computed in closed form and Mathematica came up with this:
$$\int_1^{\infty} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x}      }     \right)dx=
\frac{4}{15} \left(\sqrt{26\sqrt{2}-14 }-2\right) \approx 0.739132$$
I did a few substitutions but it didn't  turn out as an easy calculation.  I remember that eliminating square roots required some Euler type substitutions. Any ideas of how one can arrive at such a surprising result?

It's easy to show that the given integral comes down to evaluating:
$$I = \int_{0}^{1} \sqrt{\sqrt{x+1}-\sqrt{x}} \ dx$$
Define the following function on $[0,1]$:
$$f(x) = -\sqrt{\sqrt{x+1}-\sqrt{x}}$$
I claim that that this is invertible. Let's just find an explicit inverse for this. So:
$$y^2 = \sqrt{x+1}-\sqrt{x}$$
This implies that:
$$y^4+2\sqrt{x}y^2+x = x+1$$
$$\sqrt{x} = \frac{1-y^4}{2y^2}$$
which implies that $x = \frac{(1-y^4)^2}{4y^4}$. But this just means that:
$$f^{-1}(x) = \frac{1}{4x^4} -\frac{1}{2} + \frac{x^4}{4}$$
where the inverse is defined on $[-1,-\sqrt{\sqrt{2}-1}]$. Now, it follows that:
$$\int_{0}^{1} f(x) \ dx + \int_{-1}^{-\sqrt{\sqrt{2}-1}} f^{-1}(x) \ dx = -\sqrt{\sqrt{2}-1}$$
Let's evaluate the integral of the inverse function because that's easy. So, we have that:
$$\int_{-1}^{-\sqrt{\sqrt{2}-1}} f^{-1}(x) \ dx = \left[-\frac{1}{12x^3} -\frac{1}{2}x+\frac{x^5}{20}\right]_{-1}^{-\sqrt{\sqrt{2}-1}}$$
Now, I'm not going to include all the details here but you can use the above to show that:
$$I = \frac{4}{15} \sqrt{\sqrt{2}-1}(6+\sqrt{2})-\frac{8}{15}$$
This is just obtained by plugging in the bounds. But now, notice that:
$$\sqrt{\sqrt{2}-1}(6+\sqrt{2}) = \sqrt{(\sqrt{2}-1)(6+\sqrt{2})^2} = \sqrt{(\sqrt{2}-1)(38+12\sqrt{2})} = \sqrt{26\sqrt{2}-14}$$
and this gives the desired result. $\Box$
