A change of variable for $\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx$ Question 1


  I would like to understand the following change of variable


for $\phi(x)=nx,$ $\ d\phi=n\cdot dx,$ and $\chi_{E}(y)$ the usual characteristic function of a set $E$:

$$\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx= \int_0^{+\infty}\frac{\phi(x)}{n^{2/3}+n^{1/6}\phi(x)}\chi_{(0,n)}(\phi(x)) \ d\phi$$

Question 2

Is there a more slick way to evaluate the integral 

$$\lim_{n\to +\infty}\int_0^1 \frac{n^{4/3}x}{1+n^{5/2}x^3} \ dx$$
I can answer your question 1. (I don't understand your question 2, in a similar vein to @joriki's comment, and I don't understand your reply to that comment. Maybe if you elaobrate further in the comment, I will understand.)

If you make the substitution $u=nx$, then you would have
$$\int_0^1\frac{n^{4/3}x}{1+n^{5/2}x^3}dx=\int_{u=0}^{u=n}\frac{n^{1/3}u}{1+n^{-1/2}u^3}\frac{du}{n}=\int_{0}^{n}\frac{u}{n^{2/3}+n^{1/6}u^3}du$$

Now introduce that function $\chi$. Its output is $1$ if the input is in $[0,n]$, and $0$ otherwise.

$$\int_{0}^{n}\frac{u}{n^{2/3}+n^{1/6}u^3}\chi(u)\,du$$

That's fine because we just multiplied by $1$. But also we can extend the upper limit of the integral to $\infty$ because over $(n,\infty)$ this would be multiplying by $0$, and therefore not changing the value of the integral.

$$\int_{u=0}^{u=\infty}\frac{u}{n^{2/3}+n^{1/6}u^3}\chi(u)\,du$$

The posted version uses $\phi(x)$ in place of $u$, which is fine. Except I wonder if you left out an exponent of $3$ in the denominator when typing up the question.
As you've noted in the comments, this can be easily solved with DCT. However, to follow what you've tried to gone through with, we could substitute $u=nx$ to get $x=u/n$ and $\mathrm dx=\mathrm du/n$, which give

$$\int_0^1\frac{n^{4/3}x}{1+n^{5/2}x^3}~\mathrm dx=\int_0^n\frac{n^{4/3}n^{-1}u}{1+n^{5/2}(n^{-1}u)^3}~\frac{\mathrm du}n$$

Expanding out the denominator:

$$1+n^{5/2}(n^{-1}u)^3=1+n^{5/2}n^{-3}u^3=1+n^{-1/2}u^3$$

and the numerator:

$$n^{4/3}n^{-1}u=n^{1/3}u$$

which, put together and dividing by $n$ becomes

$$\int_0^n\frac{n^{4/3}n^{-1}u}{1+n^{5/2}(n^{-1}u)^3}~\frac{\mathrm du}n=\int_0^n\frac{u}{n^{2/3}+n^{1/6}u^3}~\mathrm du$$

or equivalently,

$$\int_0^n\frac{u}{n^{2/3}+n^{1/6}u^3}~\mathrm du=\int_0^\infty\frac{u}{n^{2/3}+n^{1/6}u^3}\chi_{(0,n)}(u)~\mathrm du$$

We can factor $n^{1/6}$ out from the denominator and bound the integral by

$$\int_0^\infty\frac{u}{n^{2/3}+n^{1/6}u^3}\chi_{(0,n)}(u)~\mathrm du\le\int_0^\infty\frac{n^{-1/6}u}{1+u^3}~\mathrm du$$

where $\displaystyle\int_0^\infty\frac{u~\mathrm du}{1+u^3}$ converges by seeing it is bounded on $[0,1]$ and comparing the rest to $\displaystyle\int_1^\infty\frac{u~\mathrm du}{u^3}$.