TITLE: sum, integral of certain functions
QUESTION [5 upvotes]: While working on some research, I have encountered an infinite series and its improper integral analogue:
\begin{align}\sum_{m=1}^{\infty}\frac1{\sqrt{m(m+1)(m+2)+\sqrt{m^3(m+2)^3}}}&=\frac12+\frac1{\sqrt{2}}, \\
\int_0^{\infty}\frac{dx}{\sqrt{x(x+1)(x+2)+\sqrt{x^3(x+2)^3}}}&=2.\end{align}
The evaluations were guessed using numerical evidence.
Can you provide proofs, or any reference (if available)?

REPLY [17 votes]: For the integral, notice that the expression under the square root is
  $$ x(x+1)(x+2)+x(x+2)\sqrt{x(x+2)} = \frac12\,x(x+2)(\sqrt x+\sqrt{x+2})^2. $$
Consequently,
\begin{align*} 
  \frac1{\sqrt{x(x+1)(x+2)+x(x+2)\sqrt{x(x+2)}}} 
       &= \frac{\sqrt 2}{(\sqrt x+\sqrt{x+2}) \sqrt{x(x+2)}} \\
       &= \frac1{\sqrt 2}\,\frac{\sqrt{x+2}-\sqrt{x}}{\sqrt{x(x+2)}} \\
       &= \frac1{\sqrt 2} \left( \frac1{\sqrt x}-\frac1{\sqrt{x+2}}\right);
\end{align*}
thus, the indefinite integral is
  $$ \sqrt{2}\, (\sqrt x-\sqrt{x+2})+C $$
and the result follows easily.
As Antony Quas noticed, this also works for the sum showing that the partial sum over $m\in[1,M]$ is
  $$ \frac1{\sqrt 2} \sum_{m=1}^M \frac1{\sqrt m} - \frac1{\sqrt 2} \sum_{m=3}^{M+2} \frac1{\sqrt m} = \frac1{\sqrt 2} \left( 1+\frac1{\sqrt 2}\right) + o(1). $$