<s>[INST] convergence of $\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $ For which value $\alpha$ does the following integral converges: 
$\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $

I tried to use: $ \sqrt{x^2 + 2x + 2} - 1 = \frac{x^2 + 2x + 1}{\sqrt{x^2 + 2x + 2} + 1} $
I wanted to simplify it more but that didn't lead to anything.
Thank you for your help. 
 [/INST] $I_a =\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^a dx
$
Use the standard method of
$a-b
=\dfrac{a^2-b^2}{a+b}
$.
$\begin{array}\\
\sqrt{x^2 + 2x + 2} -( x + 1)
&=\dfrac{(x^2 + 2x + 2)-(x+1)^2}{\sqrt{x^2 + 2x + 2} + x + 1}\\
&=\dfrac{(x^2 + 2x + 2)-(x^2+2x+1)}{\sqrt{x^2 + 2x + 2} + x + 1}\\
&=\dfrac{1}{\sqrt{x^2 + 2x + 2} + x + 1}\\
\end{array}
$
Therefore
$\dfrac1{2(x+1)}
\gt \sqrt{x^2 + 2x + 2} -( x + 1)
\gt \dfrac1{2(x+2)}
$.
Therefore
$I_a$ converges
if and only if
$\int_0^{\infty} \dfrac{dx}{(x+1)^a}
$
converges
and this is when
$a > 1$.
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