Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to find out for which $p \in \mathbb{R}_{>0}$ the integral

$$\int_{[0,1]^n} \frac{\mathrm d x}{(x_1^p+2x_2^p + ... + nx_n^p)^{1/3}}$$

converges. To be honest, I have no idea or whatsoever how to do this. I computed the integral for $n=1$ and $n=2$ and found them to be existing for $p < 3n$ so far. But Mathematica takes forever to compute the integral for $n=3$, so I think I have to find a smart way to verify whether my guess is correct.

Can anyone give me a little hint, so that I can proceed? The problem is that we've never done such things before during classes or in exercises, so I really don't know where to begin.

Thanks in advance.

share|improve this question
    
The inequality here should help: en.wikipedia.org/wiki/Generalized_mean#Geometric_mean –  Emre Apr 23 '11 at 18:58
    
I'm sorry for being such dull but I don't see how it helps. –  Huy Apr 23 '11 at 19:41

2 Answers 2

up vote 11 down vote accepted

This is to advocate an easy method leading to the answer without lingering on irrelevant details of the question. First, replacing each $x_i^p$ by $|x_i|^p$, one sees that it is equivalent to ask for the behaviour of the integral over $[-1,1]^n$, that is, for the behaviour of the integral around the point $x=0$. Second, since all the norms on $\mathbb{R}^n$ are equivalent, one can replace the denominator by $\|x\|^{p/3}$ for every norm $\|\ \|$, without changing the convergence or divergence. Choose the Euclidean norm and write $\|x\|=r$. By any spherical change of variables, $\mathrm{d}x$ is $r^{n-1}\mathrm{d}r$ times an infinitesimal element which involves the spherical coordinates of $x/\|x\|$. This reduces the question to the behaviour at $r=0^+$ of the integral $$ \int_0 \frac{r^{n-1}\mathrm{d}r}{r^{p/3}}. $$ Hence, there is convergence if and only if $n-1-p/3>-1$, that is, if and only if $p<3n$ (as the OP noticed when $n=1$ and $n=2$).

share|improve this answer
1  
+1, very nice answer. –  Sam Apr 24 '11 at 1:00

One upper bound on $p$ follows from this inequality:

$$\prod_{i=1}^nx_i^{w_i} \leq \sqrt[p]{\sum_{i=1}^nw_ix_i^p} for \sum w_i = 1.$$

Let $w_i = i/c$ and $c=n(n+1)/2$ and we have

$$\prod_{i=1}^nx_i^{i/c} \leq \sqrt[p]{\sum_{i=1}^n (i/c) x_i^p}$$

and

$$\frac{1}{ \sqrt[p]{\sum_{i=1}^n i x_i^p} } \leq \frac{1}{ \sqrt[p]{c} \prod_{i=1}^nx_i^{i/c} } \equiv \frac{1}{\sqrt[p]{c} } \prod_{i=1}^n x_i^{-i/c}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.