Calculate the Lebesgue integral of,
$$\lim_{n\to\infty}\int_{[0,1]}\frac{n\sqrt{x}}{1+n^2x^2}$$
I know I should use the Lebesgue dominated convergence theorem but what should be the dominating function? Can anyone give me a hint?
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1 Answer
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By AM-GM inequality: $\dfrac{n\sqrt{x}}{1 + n^2x^2} \leq \dfrac{n\sqrt{x}}{2nx} = \dfrac{1}{2\sqrt{x}}$. The dominating function is: $g(x) = \dfrac{1}{2\sqrt{x}}$
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$\begingroup$ But why is $g$ integrable on [0,1]? It is discontinuous at 0 right? $\endgroup$ May 28, 2014 at 11:35
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$\begingroup$ only at 1 point won't affect the answer, plus you integrate g it is finite. check it. $\endgroup$– DeepSeaMay 28, 2014 at 11:39
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$\begingroup$ x=0 is an integrable singularity. Note that the primitive of 1/√x is √x. $\endgroup$– UrgjeMay 28, 2014 at 11:39
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$\begingroup$ Is it because $g$ is continuous almost everywhere so it is Riemann integrable? And we haven't studied integrable singularities. $\endgroup$ May 28, 2014 at 11:45
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