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?
 A: This is a partial answer.
\begin{align}
a_n&=\int_n^{n+1} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x}}     \right)\mathbb dx\\
&=\int_n^{n+1} \sqrt{\sqrt{x}-\sqrt{x-1}}\,\mathbb dx-\int_{n+1}^{n+2} \sqrt{\sqrt{x}-\sqrt{x-1}}\mathbb\,dx\\
\sum_{n=1}^{\infty}a_n&=\int_1^{2} \sqrt{\sqrt{x}-\sqrt{x-1}}\,\mathbb dx
=\int_1^{2} \frac1{\sqrt{\sqrt{x}+\sqrt{x-1}}}\,\mathbb dx
\end{align}
The WolframAlpha shows that the indefinite integral is
$$\int \frac1{\sqrt{\sqrt{x}+\sqrt{x-1}}}\,\mathbb dx=\frac{5\left(\sqrt{x-1}+\sqrt x\right)^4+3}{15\left(\sqrt{x-1}+\sqrt x\right)^{5/2}}+C$$
, which I didn't figure out how to get there.

I completed the answer below.
\begin{align}
\sqrt{x}+\sqrt{x-1}&\to t\\
dt&=\frac12\left(\frac1{\sqrt{x}}+\frac1{\sqrt{x-1}}\right)dx=\frac12\frac{t}{\sqrt{x}\sqrt{x-1}}dx\\
t^2&=2x-1+2\sqrt x\sqrt{x-1}\\
4x(x-1)&=(t^2-2x+1)^2=t^4-2(2x-1)t^2+(2x-1)^2\\
0&=t^4-4xt^2+2t^2+1\\
x&=\frac{(t^2+1)^2}{4t^2}\\
x-1&=\frac{(t^2-1)^2}{4t^2}\\
\int \frac1{\sqrt{t}}\,\mathbb dx&=\int \frac{2\sqrt{x}\sqrt{x-1}}{t^{3/2}}\,\mathbb dt=\int \dfrac{2\dfrac{t^4-1}{4t^2}}{t^{3/2}}\mathbb dt\\
&=\frac12 \int \left(t^{1/2}-t^{-7/2}\right)\mathbb dt\\
&=\frac13t^{3/2}+\frac15t^{-5/2}+C\\
&=\frac13\left(\sqrt{x}+\sqrt{x-1}\right)^{3/2}+\frac15\left(\sqrt{x}+\sqrt{x-1}\right)^{-5/2}+C
\end{align}
A: The integral is elementary, and the antiderivative on the interval $[1, \infty)$ is real and continuous. It's honestly not hard to analytically solve it using Calc II methods and FTC II. Just bashy asf and tedious.
Let $$I(x)=\int\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x}}\text{ d}x$$
We want(excuse my notation) $$I=I(\infty)-I(1)=\int^{\infty}_1\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x}}\text{ d}x$$
Split this integral into two $$\underbrace{\int\sqrt{\sqrt{x}-\sqrt{x-1}} \text{ d}x}_A-\underbrace{\int\sqrt{\sqrt{x+1}-\sqrt{x}}\text{ d}x}_B$$
Start with A. We want to get rid of nested square roots so
$$u=\sqrt{x}$$
$$\int\sqrt{\sqrt{x}-\sqrt{x-1}} \text{ d}x=2\int u\sqrt{u-\sqrt{u^2-1}}\text{ d}u$$
Now that we have a variable outside we can sub the entire inside to get
$$v=u-\sqrt{u^2-1}$$
$$\frac12\int\frac{v^4-1}{v^{\frac{5}{2}}}\text{ d}v=\frac{4\sqrt{\sqrt x-\sqrt{x-1}}\left(4x+\sqrt{x-1}\sqrt x-2\right)}{15}$$
Repeat the same for the second integral B with the following two substitutions and simplify $$u=\sqrt{x},\qquad v=\sqrt{u^2+1}-u$$
to find that $$\int\sqrt{\sqrt{x+1}-\sqrt x}\text{ d}x=\frac{4\sqrt{\sqrt{x+1}-\sqrt{x}}\left(4x+\sqrt{x-1}\sqrt x+2\right)}{15}$$
So $$I(x)=\frac{4\sqrt{\sqrt x-\sqrt{x-1}}\left(4x+\sqrt{x-1}\sqrt x-2\right)}{15}-\frac{4\sqrt{\sqrt{x+1}-\sqrt{x}}\left(4x+\sqrt{x-1}\sqrt x+2\right)}{15}+C$$
Now plug in bounds, we see after a bit of simplification that $$I(1)=\frac{8}{15}-\frac{4}{15}(6+\sqrt2)\sqrt{-1+\sqrt2}$$
And the upper bound can be found by taking the (tedious) limit $$\lim_{x\to\infty}I(x)=0$$
(honestly for this limit you probably also can infer its value by inspecting $I(x)$ graphically or in sections to infer its asymptotic growth etc)
So $$I=-I(1)=\frac{4}{15}(6+\sqrt2)\sqrt{-1+\sqrt2}-\frac{8}{15}\approx0.739132...$$
A: There's a shorter method I thought was interesting, but it's pretty simple and it's similar to Abhijeet's answer, so feel free to downvote it if it's not useful. Going off his answer, we can simplify the integral into
$$\int_{0}^{1}\sqrt{\sqrt{x+1}-\sqrt{x}}dx.$$
Let $u = \sqrt{\sqrt{x+1}-\sqrt{x}}$ so that $x = \frac{\left(u^{4}-1\right)^{2}}{4u^{4}}$ and $dx\ =\ \frac{u^{8}-1}{u^{5}}du$. Since $\frac{\left(u^{4}-1\right)^{2}}{4u^{4}}$ is defined for all $u \in \mathbb{R}\backslash\left\{0\right\}$, we can choose our domain to be $\left(0,\infty\right)$.
When we plug in the lower bound in our function $u$, we'll get $u = 1$. When we plug in the upper bound, we'll get $u = \sqrt{\sqrt{2}-1}$.
Then the integral becomes
$$\int_{1}^{\sqrt{\sqrt{2}-1}}\frac{u^{9}-u}{u^{5}}du.$$
Just do some Reverse Power Rule stuff and you'll get the desired answer.
A: I know there already a few really good answers here on how to compute it, but I thought I could share the one I came up with as well, as there are a bunch of ways to tackle this integral. So notice first that we have that
$$I=\int_1^\infty \left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)~\mathrm{d}x=\int_0^1\sqrt{\sqrt{x+1}-\sqrt{x}}~\mathrm{d}x$$
(I could elaborate on this further if you would like, as you have to be a bit careful when proving it as you cannot just split the integral up into two due to them diverging separately). Now make the substitution
$$x=\sinh^2 u, \quad \mathrm{d}x=\sinh2u~\mathrm{d}u.$$
Using some hyperbolic identites, the fact that $\operatorname{arsinh}1=\ln(1+\sqrt{2})$ and $\operatorname{arsinh 0}=0$, and the fact that $\cosh$ and $\sinh$ are nonnegative on $[0,\ln(1+\sqrt{2})]$, we get that
$$I=\int_0^{\ln(1+\sqrt{2})}\sqrt{\sqrt{\sinh^2 u+1}-\sqrt{\sinh^2 u}}\sinh2u~\mathrm{d}u=\int_0^{\ln(1+\sqrt{2})}\sqrt{\cosh u-\sinh u}\sinh2u~\mathrm{d}u=\int_0^{\ln(1+\sqrt{2})}e^{-\frac{u}{2}}\sinh2u~\mathrm{d}u=\frac{1}{2}\int_0^{\ln(1+\sqrt{2})}e^{-\frac{u}{2}}(e^{2u}-e^{-2u})~\mathrm{d}u.$$
But now you just have to integrate a sum of two exponential functions, which is really easy to do, so I'll leave the rest to you.
A: To compute \begin{align}I=\int_{0}^{1}f(x)\;dx\end{align} with $f(x)=\sqrt{\sqrt{x+1}-\sqrt{x}}$  you can use change of variables: $x=\cot^2(u)$ then $I=\int_{\pi/8}^{\pi/4} g(u) du$ where \begin{align}g(u)=2\cot(2u)\left(2+2 \cot^2 ( 2u)   \right) \sqrt {{\frac {1-\cos(2u)}{\sin \left( 2\,u \right) }}}=4{\frac {\cos \left( 2\,u \right) }{ \sin^3(2u)}\sqrt {{\frac {\sin \left( u \right) }{\cos
 \left( u \right) }}}}
\end{align}.
Now another change of variables, $u=\arctan(v^2)$ makes
\begin{align}
I=\int_{\sqrt{\sqrt2 -1}}^{1} \frac{(1-v^4)(1+v^4)}{v^4}  \;dv
=-{\frac{8}{15}}+{\frac { \left( 4\,\sqrt {2}+24 \right) \sqrt {\sqrt {
2}-1}}{15}}
\end{align}
since $\tan\frac\pi 8=\tan\left(\frac 12 \frac \pi4\right)=\sqrt2 -1$.
A: 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. $\Box$
A: The key is that the integral has the form $i=\int_{1}^{\infty}(f(x)-f(x+1))\;dx$ which after decomposing the integration region into intervals of length 1 telescopes, so that the integral reduces to
$$\begin{align}i=\int_{0}^{1}f(x)\;dx\\\text{with }f(x)=\sqrt{\sqrt{x+1}-\sqrt{x}}\end{align}\tag{1}$$
Now we need a sequence of substitutions. It turns out to appear rather naturally in two steps. The integration then is elemntary.
The first substitution is
$$x\to z^2\tag{2}, z\in [0,1] $$
giving an integrand $2 z \sqrt{\sqrt{z^2+1}-z}$.
We now recognize the relation $\sinh^2(x)+1=\cosh^2(x)$ which suggest the substitution
$$z \to \sinh(t), t \in [0,\text{arcsinh}(1)] \tag{3}$$
leading to the integrand $\sinh (2 t) \sqrt{\cosh (t)-\sinh (t)}$, or, in exponentials, $\frac{1}{2} e^{-2 t} \sqrt{e^{-t}} \left(e^{4 t}-1\right)$. This can be integrated elementarily to give
$$i = \frac{1}{15} e^{-\frac{1}{2} \left(5 \sinh ^{-1}(1)\right)} \left(3+5 e^{4 \sinh ^{-1}(1)}\right)-\frac{8}{15}\tag{4}$$
Observing that $\sinh ^{-1}(1) = \log(1+\sqrt{2})$ gives finally
$$i= \frac{4}{15} \left(\sqrt{26 \sqrt{2}-14}-2\right)\tag{5}$$
as requested.
