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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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: $g(x) = \dfrac{1}{2\sqrt{x}}$
DeepSea
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But why is $g$ integrable on [0,1]? It is discontinuous at 0 right? – pseudogeek May 28 '14 at 11:35
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only at 1 point won't affect the answer, plus you integrate g it is finite. check it. – DeepSea May 28 '14 at 11:39
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x=0 is an integrable singularity. Note that the primitive of 1/√x is √x. – Urgje May 28 '14 at 11:39
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Is it because $g$ is continuous almost everywhere so it is Riemann integrable? And we haven't studied integrable singularities. – pseudogeek May 28 '14 at 11:45
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No. I get it now. Thanks. – pseudogeek May 28 '14 at 12:02
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If AM-GM isn't part of your box of tools you can also write $$\frac{n \sqrt x}{1 + n^2 x^2} = \frac{ nx}{ 1 + n^2 x^2} \frac{1}{\sqrt{x}}$$ and use the fact that $\phi(t) = \dfrac{t}{1 + t^2}$ is bounded on $[0,\infty)$. – Umberto P. May 28 '14 at 19:21