# Is there an integrable dominating function?

I want to justify taking the limit under the integral for

$$\lim_{n \to \infty}\int_0^1 \frac{n^2x}{1+n^3x^2}dx = \int_0^1 \lim_{n \to \infty}\frac{n^2x}{1+n^3x^2}dx = 0.$$

I know the limit is $0$ from evaluating the integral. But can I use a theorem such as monotone or dominated convergence?

The convergence is not bounded or monotone, so I tried the uniform and dominated convergence theorems. Looking at the maximum of the integrand I see it growing like $n$ so I don’t have uniform convergence.

But I also can't find a dominating function that works. Is there some standard approach I can use?

Thank you for any suggestions.

For fixed $x \in (0,1]$, the maximum of $f(n,x) = n^2 x/(1+n^3 x^2)$ occurs at $n = 2^{1/3}/x^{2/3}$, with $f(2^{1/3}/x^{2/3}, x) = 2^{2/3}/(3 x^{1/3})$. This is integrable on $[0,1]$. Thus you can use Dominated Convergence.

The function $\displaystyle f_n(x) = \frac{n^2x}{1 + n^3x^2}$ attains a maximum value of $\displaystyle \frac{n^{1/2}}{2}$ on $[0,1]$ at $x = n^{-3/2}$. A good guess would be a dominating function of the form $g(x) = Cx^{-\alpha}$ -- with $\alpha <1$ in order that $g$ be integrable.

At the maximum we can enforce

$$\frac{n^{1/2}}{2} \leqslant g(x^{-3/2}) = Cx^{3\alpha/2},$$

with $C > 1/2$ and $\alpha = 1/3$.

We then have $f_n(1) \leqslant 1/2 < g(1)$ and $f_n(x) < g(x)$ for $0 < x < n^{-3/2}$ since $f_n(x)$ decreases to $0$ and $g(x)$ increases to $\infty$ as $x \to 0.$

We should be able to find a sufficiently large $C$ to ensure $g(x) > f_n(x)$ on the interval $(n^{-3/2},1)$ as well.

Specifically, $$\frac{n^2x}{1+ n^3x^2} \leqslant Cx^{-1/3} \implies h(x) = \frac{n^2x^{4/3}}{1+ n^3x^2} \leqslant C.$$

Setting $h'(x) = 0$. we find a maximum at $x = \sqrt{2/n^3}$ where $h(\sqrt{2/n^3})= 4^{1/3}/3.$

Thus, an integrable dominating function is

$$g(x) = \frac{4^{1/3}}{3} x^{-1/3}$$

In this case, you can get away by simply calculating the integral.

Let $y= n^2 x^2$. Then $dy =2 n^2 x dx$, so that

$$I = \frac 12 \int_0^{n^2} \frac{dy}{1+n y}=\frac{1}{n} \ln (1+ny)|_0^{n^2}=\frac{\ln (1+n^3)}{n}.$$