# Evaluate the integral: $\lim \limits_{n\to\infty}\int_0^1\frac{nx}{nx^3+1}$

Evaluate the integral: $$\lim \limits_{n\to\infty}\int_0^1\frac{nx}{nx^3+1}dx$$

I'm pretty much stuck on how to solve this one: $$\int_0^1\frac{nx}{nx^3+1}dx$$

or even getting the improper integral.

What can i do?

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Do you know about the Dominated Convergence Theorem? Your problem is a simple application of it. –  nayrb Jul 10 '13 at 18:04

## 5 Answers

If you think that your integral is $\int_0^1\frac{x}{x^3+1/n}$, in the limit it is the integral of $1/x^2$, so you should expect it to diverge.

Then you can do the following: $$\int_0^1\frac{nx}{nx^3+1}=\int_0^1\frac{x}{x^3+1/n}\geq\int_{1/n^{1/3}}^1\frac{x}{x^3+1/n}\geq\int_{1/n^{1/3}}^1\frac{x}{2x^3}\\ \ \\=\int_{1/n^{1/3}}^1\frac1{x^2}=n^{1/3}-1.$$ So $$\lim_n\int_0^1\frac{nx}{nx^3+1}=\infty.$$

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No $dx$? It's all wrong! –  Git Gud Jul 10 '13 at 18:10
I'll put the $dx$ if it makes you happy... ;) –  Martin Argerami Jul 10 '13 at 18:10
$\ddot \smile{{}}$ –  Git Gud Jul 10 '13 at 18:11
I think i got it, Thanks. –  StationaryTraveller Jul 10 '13 at 18:13

Not as good, as other answers, but still...

Since $\ 0<x<1$$\int\limits_0^1\frac{nx}{nx^3+1}dx>\int\limits_0^1\frac{nx}{nx^2+1}dx=\ln\sqrt{n+1}\to\infty$$ - Actually, this one is great. – StationaryTraveller Jul 10 '13 at 19:31 I like this one better than mine. – Martin Argerami Jul 11 '13 at 4:26 Thanks. Others are just more obvious and give general idea of how to handle these types of problems, so they are more useful. Mine is not so obvious. But I can see why you like mine more - it's somewhat shorter and simplier ) Sorry for possible language mistakes, I'm Russian. – George Stobbart Jul 11 '13 at 9:22 Step 1: $$\frac{nx}{nx^3+1}=\frac{x}{x^3+\frac{1}{n}}$$ Step 2: $$\int_0^1 \frac{x}{x^3}dx~~\textrm{ diverges}$$ - so here do we need to show uniform Convergence and then take limits inside? – Neeraj Bhauryal Jul 10 '13 at 18:08 So you first calculated the limit? Is it mathematically correct? – StationaryTraveller Jul 10 '13 at 18:08 This isn't a complete solution, just a pair of hints. See Martin's answer for a complete treatment. – vadim123 Jul 10 '13 at 18:09 I got it, Thank you. – StationaryTraveller Jul 10 '13 at 18:13 To add it in more explicit language: You would like to exchange the limit and the integral, i.e. to write $$\lim_{n \to \infty} \int_0^1 f_n(x) dx = \int_0^1 \lim_{n \to \infty} f_n(x) dx = \int_0^1 \frac{dx}{x^2} = +\infty.$$ The Lebesgue Monotone Convergence Theorem allows to do just that. - @MartinArgerami sorry, my bad. Will correct in a sed. – gt6989b Jul 10 '13 at 18:14 I'd try to split this integral using partial fractions, treating$n\$ as a constant :

$$\begin{eqnarray} I &=& \int_0^1 \frac{nx}{n x^3 + 1} dx \\ &=& \int_0^1 \frac A{n^{1/3} x + 1} dx + \int_0^1 \frac {B x + C}{n^{2/3} x^2 - n^{1/3}x + 1} dx \end{eqnarray}$$

Both integrals can be done using classical techniques. The rest is up to you.

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