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I have the following integral: $$\lim_{n \to \infty} \int_{0}^{1} \frac{n\sqrt{n}x}{1+n^2x^2} \, \mathrm{d}x $$ To use the dominated convergence theorem I know that limit of $f_n$ is $0$ and $|f_n|<\frac{n^{1/2}}{2}$. However, I am having trouble to find a function that is greater than $\frac{n^{1/2}}{2}$ for all n. Can someone help? Thanks in advance.

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    $\begingroup$ Why don't you just evaluate the integral? $\endgroup$ – Botond Nov 6 '19 at 18:17
  • $\begingroup$ I need to use lebesgue integration techniques to solve it unfortunately. $\endgroup$ – Onur Bilge Nov 6 '19 at 18:20
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    $\begingroup$ The technique you are using here is that the Lebesgue integral equals the Riemann integral if both exist ;) $\endgroup$ – Maximilian Janisch Nov 6 '19 at 18:20
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For fixed $x \in (0,1]$ the maximum of $\frac{n^{3/2}x}{1 + n^2x^2}$ is attained at $n = \frac{3^{1/2}}{x}$, providing an integrable dominating function:

$$\frac{n^{3/2}x}{1 + n^2x^2} \leqslant \frac{3^{3/4}}{4} \frac{1}{\sqrt{x}}$$

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  • $\begingroup$ But how do you find at which $n$ $f$ attains maximum? $\endgroup$ – Onur Bilge Nov 6 '19 at 19:01
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    $\begingroup$ Maximize using calculus -- set derivative with respect to $n$ equal to $0$. $\endgroup$ – RRL Nov 6 '19 at 19:02
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    $\begingroup$ It is a little bit easier to do the maximization by writing $\frac{n^{3/2} x}{1+n^2 x^2}=\left ( n^{-3/2} x^{-1} + n^{1/2} x \right )^{-1}$ and then you have to minimize $n^{-3/2} x^{-1} + n^{1/2} x$. This is a bit more easily done by setting $m=n^{1/2}$ so you have to minimize $m^{-3} x^{-1} + m x$ which gives the equation $-3m^{-4} x^{-1} + x = 0$. $\endgroup$ – Ian Nov 6 '19 at 19:14
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As pointed out by Botond, evaluating the integral readily leads to $$ \frac{1}{2\sqrt{n}} \, \ln \left( 1+n^2\right) \, $$ the limit of which is clearly $0$ as $n \to\infty$.

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There is no function greater than $\frac{n^{1/2}}{2}$ for all $n$. So you will need a better upper bound. Your upper bound may depend on $x$, but not on $n$. Here, we will need an upper bound that ${}\to \infty$ as $x \to 0$.


For example, it is enough to show $|f_n(x)| \le x^{-1/2}$ for all $n$ and all $x$, since $\int_0^1 x^{-1/2} dx$ converges.

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