# Show that $\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty$

Here's my solution to an old qualifier problem. Would you tackle it differently? Is there a flaw in my work?

Suppose that $\alpha_1, \dotsc, \alpha_n$ are positive numbers such that $$\frac1{\alpha_1}+ \dotsb + \frac1{\alpha_n}<1.$$

Show that $$\int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}}<\infty.$$

My answer: $x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n} \overset{\text{(AM$\geq$GM)}}{\geq} n\cdot \sqrt[n]{x_1^{\alpha_1}\dotsb x_n^{\alpha_n}}$

$$\Rightarrow \int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1}+\dotsb + x_n^{\alpha_n}} \leq \frac1n \int_{1}^\infty \dotsb\int_{1}^\infty \frac{dx_1 \dotsb dx_n}{x_1^{\alpha_1/n}\dotsb x_n^{\alpha_n/n}},$$

at which point if we switch to hyperspherical coordinates

$$\leq \frac1n \int_{S^{n-1}} \int_{1}^\infty \frac{dx_1 \dotsb dx_n}{r^{\frac{\alpha_1+\dotsb + \alpha_n}{n}}|f(\sigma)|}r^{n-1} dr\,d\sigma,\tag{1}$$

where $\sigma$ is the position on the unit $(n-1)$ sphere in $\mathbb{R}^n$ and $|f(\sigma)|\leq1$. Note I have chosen a crude bound by integrating over the entire area outside the unit ball. Now

$$\frac1{\alpha_1}+ \dotsb + \frac1{\alpha_n}<1 \\\Rightarrow \frac{n}{\frac1{\alpha_1}+ \dotsb + \frac1{\alpha_n}}>n\\\overset{\text{AM \geq HM}}{\Rightarrow}\frac{\alpha_1 + \dotsb+\alpha_n}{n}>n$$

so the integral (1) is integrating a power of $r$ which is $>1$ in the denominator, so it converges.

• By Fubini's theorem, the integral related to the GM will diverge when $\alpha_i\le n$ for some $i$. – 23rd Oct 17 '13 at 20:06
• Shoot, you're right! Where is the flaw in my logic? – Eric Auld Oct 17 '13 at 20:10
• The flaw is you didn't obtain any control of the integral of $\frac{1}{|f(\sigma)|}$ over $S^{n-1}$. – 23rd Oct 17 '13 at 21:00
• Of course! ---- – Eric Auld Oct 18 '13 at 2:05

The AM-GM could be modified as follows.

Denote $$p:=\left(\frac{1}{\alpha_1}+\cdots+\frac{1}{\alpha_n}\right)^{-1}>1,$$ and denote $$y_i=\alpha_i\log x_i,\quad\lambda_i=\frac{p}{\alpha_i}\in (0,1), \quad i=1,\cdots,n.$$ Since $\sum_{i=1}^n\lambda_i=1$, by Jensen's inequality for the exponential function,

$$\sum_{i=1}^n x_i^{\alpha_i}\ge \sum_{i=1}^n \lambda_i e^{y_i}\ge \exp\left(\sum_{i=1}^n \lambda_i y_i\right)=\prod_{i=1}^n x_i^p.\tag{1}$$

By $(1)$ and Fubini's theorem, the original integral is bounded by $$\left(\int_1^\infty\frac{dx}{x^p}\right)^n=\frac{1}{(p-1)^n}.$$

• Wow, that is a very useful result. Thanks! – Eric Auld Oct 18 '13 at 14:23
• @EricAuld: You are welcome! – 23rd Oct 18 '13 at 15:59
• Where can I find out more about methods like the one above for $\sum_1^n x_i^{\alpha_i} \geq \prod_1^nx_i^p$? I don't think I would have come up with that on my own. – Eric Auld Oct 18 '13 at 16:38
• @EricAuld: I don't know, but if you are familiar with Jensen's inequality, then you can apply it to different convex functions to produce kinds of inequalities like the one you mentioned as you wish. – 23rd Oct 18 '13 at 16:55

As already said on the page, the AM-GM inequality yields a sufficient condition too strong to be necessary. Instead one can rely directly on the change of variable $x_k^{\alpha_k}=ru_k^2$ with $r$ in $(1,\infty)$ and $u=(u_k)$ in the unit sphere $\sum\limits_ku_k^2=1$. Then the Jacobian of the transformation $(x_k)\to(r,u)$ is complicated to write down completely but it has the form $\prod\limits_k\mathrm dx_k=r^{\beta-1}\mathrm dr\mathrm d\mu(u)$ with $\beta=\sum\limits_k\frac1{\alpha_k}$, where $\mu$ is some finite measure on the unit sphere. Then $\sum\limits_kx_k^{\alpha_k}=r$ hence the integral to be considered converges if and only if $$\int^\infty\frac{r^{\beta-1}\mathrm dr}r$$ converges, that is, if and only if $\beta\lt1$.

• Thank you for helping me understand the problem better. This is very simple. Could you perhaps help me understand why $\prod_k \mathrm{d}x_k = r^{\beta - 1}dr d\mu(u)$? – Eric Auld Oct 18 '13 at 15:09
• Well, you could try to compute the Jacobian when $x_k=r^{1/\alpha_k}\varphi_k(u)$ for each $k$. – Did Oct 18 '13 at 15:18
• Ok, I get $$\frac{\partial\mathbf{x}}{\partial(r,\mathbf{u})} = \left( \begin{matrix} \frac{1}{\alpha_1}r^{1/\alpha_1 -1} & \vec{\frac{\partial x_1}{\partial \mathbf{u}}} \\ \vdots & \vdots \\ \frac{1}{\alpha_n}r^{1/\alpha_n -1} & \vec{\frac{\partial x_n}{\partial \mathbf{u}}}\end{matrix} \right),$$but how would I compute the determinant of something like that? (Or at least show it is of the form $r^{\beta -1}$? Thanks for your help – Eric Auld Oct 18 '13 at 15:33
• Actually the Jacobian matrix is an $n\times n$ matrix whose first column is what you wrote and whose other columns are all multiples of the column of $r^{1/\alpha_k}$. Hence when one develops the determinant, every product multiply every factor $r^{1/\alpha_k}$ except one which is $r^{1/\alpha_k-1}$. Thus, every term is a factor of $r^{\beta-1}$. – Did Oct 18 '13 at 21:17
• How do we know that $(x_k) \mapsto (r,u)$ is a diffeomorphism onto its image? – Eric Auld Nov 21 '13 at 21:59


\begin{align} {\cal J} &= \int_{1}^{\infty}\cdots\int_{1}^{\infty}\bracks{% \int_{0}^{\infty}\expo{-\pars{x_{1}^{\alpha_{1}} + \cdots + x_{n}^{\alpha_{n}}}\mu}\,\dd\mu} \dd x_{1}\ldots\dd x_{n} \\[3mm]&= \int_{0}^{\infty}\dd\mu\prod_{i = 1}^{n}\int_{1}^{\infty} \expo{-\pars{x^{\alpha_{i}}}\mu}\,\,\dd x = \int_{0}^{\infty}\dd\mu\prod_{i = 1}^{n} {1 \over \alpha_{i}\,\mu^{1/\alpha_{i}}} \int_{\mu}^{\infty}\expo{-x}\,x^{\pars{1/\alpha_{i}} - 1}\,\,\dd x \\[3mm]&= \int_{0}^{\infty}\prod_{i = 1}^{n} {\Gamma\pars{1/\alpha_{i}\,,\,\mu} \over \alpha_{i}\,\mu^{1/\alpha_{i}}} \,\dd\mu = \int_{0}^{\infty} {1 \over \mu^{1/\alpha_{1} + \cdots + 1/\alpha_{n}}} \prod_{i = 1}^{n} {\Gamma\pars{1/\alpha_{i}\,,\,\mu} \over \alpha_{i}\,} \,\dd\mu \end{align}

$\Gamma\pars{z,a}$ is the $\it\mbox{Incomplete Gamma function}\ \pars{~\Re\, z > 0~}$ and $\Gamma\pars{z} \equiv \Gamma\pars{z,0}$ is the $\it\mbox{Gamma function}$.

1. $\large\mu\ \gtrsim\ 0$: $${1 \over \mu^{1/\alpha_{1} + \cdots + 1/\alpha_{n}}} \prod_{i = 1}^{n} {\Gamma\pars{1/\alpha_{i}\,,\,\mu} \over \alpha_{i}\,}\ \sim\ {1 \over \mu^{\Lambda}} \prod_{i = 1}^{n} {\Gamma\pars{1/\alpha_{i}} \over \alpha_{i}\,}\,, \quad \Lambda \equiv {1 \over \alpha_{1}} + \cdots + {1 \over \alpha_{n}} < 1$$ Since $0 < \Lambda < 1$, the integral ~ $\mu^{1 - \Lambda}$ when $\mu \gtrsim 0$
2. $\large \mu\ \gg\ 1:$

In this case, $\ds{% \Gamma\pars{{1 \over \alpha_{i}},\mu} \sim \mu^{\pars{1/\alpha_{i}} - 1}\,\expo{-\mu}}$ $${1 \over \mu^{1/\alpha_{1} + \cdots + 1/\alpha_{n}}} \prod_{i = 1}^{n} {\Gamma\pars{1/\alpha_{i}\,,\,\mu} \over \alpha_{i}\,}\ \sim\ {\expo{-n\mu} \over \mu^{n}} \prod_{i = 1}^{n}{1 \over \alpha_{i}\,}\,,$$

So far, we can conclude that the condition $\ds{\sum_{i = 1}{1 \over \alpha_{i}} < 1}$ guarantees the convergence of the integral.

IMHO the main flaw in your logic is that you have generated a sufficient but not a necessary condition for the original integral to converge. You write: $$\int_1^\infty \cdots \int_1^\infty \frac{dx_1 \cdots dx_n}{x_1^{\alpha_1}+\cdots + x_n^{\alpha_n}} \leq \frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty \frac{dx_1 \cdots dx_n}{x_1^{\alpha_1/n}\cdots x_n^{\alpha_n/n}}$$ But: $$\frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty \frac{dx_1 \cdots dx_n}{x_1^{\alpha_1/n}\cdots x_n^{\alpha_n/n}} = \frac{1}{n} \int_{1}^\infty \frac{dx_1}{x_1^{\alpha_1/n}} \cdots \int_{1}^\infty \frac{dx_n}{x_n^{\alpha_n/n}}$$ And: $$\int_{1}^\infty \frac{dx_i}{x_i^{\alpha_i/n}} = \left[ \frac{x_i^{1-\alpha_i/n}}{1-\alpha_i/n} \right]_1^\infty = \left\{ \begin{array}{ll} \frac{1}{1-\alpha_i/n} & \mbox{if} \quad \alpha_i > n \\ \infty & \mbox{if} \quad \alpha_i \le n \end{array} \right.$$ So instead of $1/\alpha_1 + \cdots + 1/\alpha_n < 1$ we have $(1/\alpha_1 < 1/n) \wedge \cdots \wedge (1/\alpha_n < 1/n)$, which is a much stronger condition.