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I encountered the following problem on a practice exam:

Suppose $a_1,...,a_n>0$ are such that $$\sum_{k=1}^n\frac{1}{a_k}<1.$$ Then $$\int_1^\infty\cdots\int_1^\infty\frac{1}{x_1^{a_1}+\cdots+x_n^{a_n}}\,dx_1\cdots dx_n<\infty.$$

Now, this is not tricky for $n=1$, as $\frac{1}{a_1}<1$ iff $a_1>1$ in this circumstance, and so convergence is guaranteed by the convergence of $\sum_{k=1}^\infty k^{-a_1},$ but I'm not sure how to proceed for other $n$. I have a few thoughts:

(1) Actually calculate the antiderivative at each stage, and apply methods of improper integrals where necessary. This seems like it would be more hectic than it's worth.

(2) Find an upper bounding integrand that is more friendly, and show that that integral evaluates to a real number. This one seems like it could be doable, if I can find such a bounding integrand. My attempts so far have either failed to evaluate to a real number or failed to be a bounding integrand.

(3) Generalize the integral test and the result that $\sum_{k=1}^\infty k^{-p}$ is a real number for $p>1$ in some fashion. This seems like it could be my best bet, but I worry that if I don't phrase them carefully, I may be trying to prove results that are incompatible or not both true.

My question is this: Does anyone have any recommendations for how I might proceed (whether it's a hint for one of the approaches I listed above or the seed of an entirely different approach)? Thanks for any help you can give me!

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Well, if you proceed by induction, then the integral of $1/(x_1^{a_1} + X)$ (where $X = x_2^{a_2} + \cdots + x_n^{a_n})$ can be conveniently expressed as a hypergeometric function in terms of $x$, $X$ and $a_1$. By induction hypothesis, you'll get the result (the form of this hypergeometric function is a quotient, where in the denominator you'll have $X$ and the numerator will be finite. –  William Jun 18 '12 at 19:48

1 Answer 1

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There is a method which uses spherical coordinates. Let $A_1,\ldots,A_N\geq 1$. We have to show that $$I(A_1,\ldots,A_n):=\int_{\prod_{j=1}^n[1,A_j]}\frac 1{\sum_{j=1}^nx_j^{a_j}}dx_1\ldots dx_n$$ can be bounded independently of the $A_1,\ldots, A_n$. We use the substitutions $t_j^2=x_j^{a_j}$, $x_j=t_j^{\frac 2{a_j}}$, which give $$I(A_1,\ldots,A_n)=C\int_{\prod_{j=1}^n[1,A_j^{\frac{a_j}2}]}\frac{\prod_{j=1}^nt_j^{2/a_j-1}}{|t|^2}dt_1\dots dt_n,$$ where $C$ depend only on $a_j$. We note that $\prod_{j=1}^n[1,A_j^{\frac{a_j}2}]$ is contained in a set of the form $\{x\in\Bbb R^n, r_1\leq |x|\leq r_2\}$, where $r_1>0$ (doesn't depend on the $A_j$) and $|\cdot|$ is the euclidian norm. We get $$I(A_1,\ldots,A_n)\leq C\int_{\{r_1\leq |t|\leq r_2\}}\frac{\prod_{j=1}^nt_j^{2/a_j-1}}{|t|^2}dt_1\dots dt_n.$$ Now we use spherical coordinates: the Jacobian is $r^{n-1}$ times terms in $\sin$ and $\cos$ which can be bounded by $1$. We will also bound the terms coming from the $t_j$ in the numerator, to get $$I(A_1,\ldots,A_n)\leq C'\int_{r_1}^{r_2}r^{2\sum_{j=1}^n\frac 1{a_j}}r^{-n}r^{n-1}r^{-2}dr=C'\int_{r_1}^{r_2}r^{2\sum_{j=1}^n\frac 1{a_j}-3}dr.$$ Since $2\sum_{j=1}^n\frac 1{a_j}-3<-1$, the integral $$\int_{r_1}^{+\infty}r^{2\sum_{j=1}^n\frac 1{a_j}-3}dr$$ is convergent and we are done.

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