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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.

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The inequality here should help: – 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
up vote 12 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 $n=2$).

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+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

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


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

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