# Order of an integral

Why is this integral $$\int_{2\pi}^0 {x^ae^{ia\theta}\over 1+\sqrt{2} x e^{i\theta}+x^2e^{2i\theta}}ixe^{i\theta}d\theta=O(x^{a+1})\to 0$$ as $x\to 0$,where $|a|<1$?

I understand that limit value bit (!) but I don't see why the integrand should be of that order. I would have thought that the order should be $O(x^a)$.

Source : these notes.

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Why do you think it should be $O(x^a)$? You can factor out the $x^a$ and the $x$ (did you miss the $x$ sitting there between the $i$ and the $e^{i\theta}$?) so it's $x^{a+1}$ times the integral of something not far from $1$ in modulus, so $O(x^{a+1})$.
Thanks, Gerry, My mistake was I looked at the denominator and saw that it is of order $2$ and the numerator is of order $a+1$ and cancelled them, which I really shouldn't have done! Thanks again! –  Zemlinsky Mar 6 '12 at 12:06