# Evaluating a Lebesgue Integral

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.

• Why don't you just evaluate the integral? – Botond Nov 6 '19 at 18:17
• I need to use lebesgue integration techniques to solve it unfortunately. – Onur Bilge Nov 6 '19 at 18:20
• The technique you are using here is that the Lebesgue integral equals the Riemann integral if both exist ;) – Maximilian Janisch Nov 6 '19 at 18:20

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:

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

• But how do you find at which $n$ $f$ attains maximum? – Onur Bilge Nov 6 '19 at 19:01
• Maximize using calculus -- set derivative with respect to $n$ equal to $0$. – RRL Nov 6 '19 at 19:02
• 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 $-3m^{-4} x^{-1} + x = 0$. – Ian Nov 6 '19 at 19:14

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 $$n \to\infty$$.

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.