Calculate $\lim_{n\to\infty}\int_0^1\frac{nx}{1+n^2 x^3}\,dx$.

Question

I need calculate $\displaystyle\lim_{n\to\infty}\int_0^1\frac{nx}{1+n^2 x^3}\,dx$.

I showed that $f_n(x)=\frac{nx}{1+n^2 x^3}$ converges pointwise to $f\equiv 0$ but does not converge uniformly in [0,1].

Also $f_n$ has a maximum when $\displaystyle x=\frac{1}{\sqrt[3]{2n^2}}$.

By the last thing, I can't use the theorem of integration of uniform convergence.

I wait that you can give me a hint for this exercise. Thanks!

Answer 1

By $AM-GM$, $$2n\sqrt{x^3}=2\sqrt{n^2x^3} \le 1+n^2x^3.$$

So then you can bound the integrand by $$\frac{nx}{1+n^2x^3} \le \frac{nx}{2nx^{3/2}} =\frac{1}{2\sqrt{x}}$$ which is integrable on $[0,1]$, and then apply dominated convergence.

Answer 2

Let $t=nx$ and then \begin{eqnarray} I_n&=&\int_0^1\frac{nx}{1+n^2 x^3}\,dx\\ &=&\int_0^n\frac{t}{1+n^2 \frac{t^3}{n^3}}\frac{1}{n}\,dt\\ &=&\int_0^n\frac{t}{n+t^3}dt\\ &=&\int_0^\infty\chi_{[0,n]}(t)\frac{t}{n+t^3}dt\\ \end{eqnarray} Noting $$ \chi_{[0,n]}(t)\frac{t}{n+t^3}\le\frac{t}{1+t^3}, \int_0^\infty\frac{t}{1+t^3}dt<\infty, \lim_{n\to\infty}\chi_{[0,n]}(t)\frac{t}{n+t^3}=0 $$ by the DCT, one has $$ \lim_{n\to\infty}I_n=0. $$

Answer 3

With $t:=n^{2/3}x$,$$\lim_{n\to\infty}\int_0^1\frac{nx}{1+n^2 x^3}\,dx=\lim_{n\to\infty}n^{-1/3}\int_0^{n^{2/3}}\frac t{1+t^3}\,dt$$

and the expression is asymptotic to

$$\frac{2\pi}{3\sqrt3}n^{-1/3}.$$

(Definite integral obtained with Wolfram Alpha, but the exact value is inessential.)

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We can be more accurate by evaluating the asymptotics of the integral,

$$\int_0^{n^{2/3}}\frac t{1+t^3}dt=\int_0^\infty\frac t{1+t^3}dt-\int_{n^{2/3}}^\infty\frac t{1+t^3}dt\approx\frac{2\pi}{3\sqrt3}-n^{-2/3},$$

as for large $t$, $\dfrac t{1+t^3}\approx\dfrac1{t^2}$.

Hence,

$$L_n\sim\frac{2\pi}{3\sqrt3}n^{-1/3}-\frac1n.$$