<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] $$\sqrt{x^2+2x+2}-x-1=\frac{x^2+2x+2-(x^2+2x+1)}{\sqrt{x^2+2x+2}+x+1}=\frac{1}{\sqrt{(x+1)^2+1}+x+1}\sim \frac{1}{2(x+1)}$$
So our integral converges if
$$\int_0^\infty\frac{dx}{(2(x+1))^\alpha}$$
 converges, do you know when this integral converges? You can compute the anti derivative to assure yourself.
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