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I'm trying to prove that

$\displaystyle \int_{[2,\infty]} \frac{x\,dx}{\sqrt{(x^2-\epsilon^2)(x^2-1)(x-2)}}\;\;$ converges to $\;\;\displaystyle \int_{[2,\infty]}\frac{dx}{\sqrt{(x^2-1)(x-2)}}\;\;$ as $\epsilon\rightarrow\,0$

I thought I could use Dominated Convergence Theorem, but I'm having trouble finding an integrable dominating function because of what happens at x=2. Can I use the 'almost everywhere' part to split the interval somehow?

Is the theorem applicable, or should I be thinking about this differently?

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1 Answer 1

The statement that ${x \over \sqrt{x^2 - \epsilon^2}} \leq 2$ is equivalent to $x^2 \leq 4(x^2 - \epsilon^2)$ which is the same as $4\epsilon^2 \leq 3x^2$, which will hold on your whole domain as long as $\epsilon < 1$ for example. So you can use $2{\sqrt{(x^2-1)(x-2)}}$ as your dominating function (it's integrable).

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