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Let $a_i >0$ for $1\le i \le n$, and let $J=(0,1) \times \dots \times (0,1)$.

I want to prove : $$\int_J{1 \over x_1^{a_1}+x_2^{a_2}+ \dots+x_n^{a_n}}dx<\infty \Longleftrightarrow \sum^n_{i=1} {1 \over a_i}>1$$

It's not simple problem I think. How can I prove that? Should I use Fubini theorm?

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up vote 3 down vote accepted

Here is a proof that $\sum_{i=1}^n \frac{1}{a_i}>1$ implies the integral is finite: Denote $\sum_{i=1}^n \frac{1}{a_i}$ as $\frac{1}{a}$, then $\sum_{i=1}^n \frac{a}{a_i} = 1$. There is an arithmetic-geometric mean inequality, according to which $y_1^{s_1}y_2^{s_2}\cdots y_n^{s_n}\le s_1y_1+\cdots+s_ny_n$ when the $y_i$ are positive and the sum of the $s_i$ is 1. Apply this using $y_i=x_i^{a_i}$ and $s_i = \frac{a}{a_i}$. You find that $$y_1^{a/a_1}y_2^{a/a_2}\cdots\le\frac{a}{a_1}y_1+\frac{a}{a_2}y_2+\cdots$$ The left hand side is equal to $(x_1x_2\cdots)^a$, and the right side is less than $y_1+y_2+\cdots$. Therefore $$(x_1x_2\cdots)^a\le x_1^{a_1}+x_2^{a_2}+\cdots$$ The reciprocal gives $$\frac{1}{x_1^{a_1}+\cdots}\le\frac{1}{x_1^ax_2^a\cdots}$$ If $a<1$ then each integral $\int_0^1 \frac{1}{x_i^a}\,dx_i$ converges. Therefore, if $a<1$, then the integral converges.

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Thanks for your perfect answer. – japee Jun 11 '12 at 4:32

Here's a way to see both directions roughly at the same time. For any $1\le k\le n$, let $J_k$ be the subregion $\{ x \in J : x_k^{a_k} = \max(x_1^{a_1}, \ldots, x_n^{a_n})\}$. Equivalently, $x \in J_k$ iff $x_i \le x_k^{a_k/a_i}$ for each $1 \le i \le k$.

Notice that for any $x \in J_k$ we have $$x_k^{a_k} \le \sum_{i=1}^n x_i^{a_i} \le nx_k^{a_k}.$$ Let's define $I_k = \int_{J_k} x_k^{-a_k} dx.$ By the above inequalities, we clearly have $I \ge \frac1n I_k$ since $J_k \subset J$, and also $I \le I_1 + I_2 + \cdots + I_n$, because $J_1 \cup \cdots \cup J_n = J$.

Therefore $I$ converges if and only if each $I_k$ converges. It is now a routine matter to calculate $I_k$ by iterated integration, saving $x_k$ for last. The innermost $n-1$ integrals are just the volume of a box with dimensions $x_k^{a_k/a_i}$ (from the definition of $J_k$). For instance, when $k=1$ we get

$$I_1 = \int_0^1 \left(\prod_{i=2}^n x_1^{a_1/a_i} \right) x_1^{-a_1} dx_1 = \int_0^1 x_1^{-a_1 + a_1/a - 1} dx_1,$$ where $\frac1a = \frac1{a_1} + \cdots + \frac1{a_n}$ as in @BobTerrell's solution. This one-dimensional integral converges iff $-a_1 + a_1/a > 0$, in other words exactly when $a < 1$.

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