0
$\begingroup$

I am considering the following function. $f(x)=\sum_{i=1}^d{\mid x_i\mid^2}$ for $x\in\mathbb{R}^d$.

I am now considering the inverse function $g(x)=\frac{1}{f(x)}$.

Claim: g is integrable if $d\geq3$. Does this claim is correct? It does not seems very reasonable but I tried anyway manual computation for $d\geq3$ but it seems not to be that useful. Does someone has in case a more direct way to prove this? Thanks

$\endgroup$
5
  • $\begingroup$ Polar coordinates. $\endgroup$
    – Hetebrij
    May 14, 2016 at 7:48
  • $\begingroup$ @Hetebrij Is it true then to consider the d dimensional spherical coordinate? $r^2=\sum_{i=1}^d{x_i^2}$ and $\phi_k=arccot(\frac{x_k}{\sqrt{r^2-\sum_{i=1}^k{x_i^2}}})$. Then we get the integral $\int_V{\frac{1}{r^2}(r^{n-1}\sin^{n-2}(\phi_1)\dots\sin(\phi_{n-2}))drd\phi_1... d\phi_{n-2}}$? $\endgroup$
    – sky90
    May 14, 2016 at 8:59
  • 1
    $\begingroup$ Yes, but you miss a variable. But then the integral becomes some finite factor times a simple one dimensional integral, which is only finite on bounded sets $V$. $\endgroup$
    – Hetebrij
    May 14, 2016 at 9:31
  • $\begingroup$ I see. great thanks! $\endgroup$
    – sky90
    May 14, 2016 at 9:35
  • $\begingroup$ Then I would suggest making this an answer and accepting your answer for future users :). $\endgroup$
    – Hetebrij
    May 14, 2016 at 9:36

1 Answer 1

Reset to default
0
$\begingroup$

Let the variables $r^2=\sum_{i=1}^d{x_i^2}$, $\phi_k=\text{arccot}\left[\frac{x_k}{\sqrt{r^2-\sum_{i=1}^k{x_i^2}}}\right]$ and since every $x_k\in[-\pi,\pi]$ we have that $\phi_k\in [-\pi/2,\pi/2]$ and $r\in[0,\pi\sqrt{d}]$. Therefore using this transformation we get the following. \begin{equation} \begin{split} \int_{[-\pi,\pi]^d}{\mid\theta\mid^{-2}d\theta}&=\int_{[-\pi/2,\pi/2]^{d-1}}{\int_0^{\pi\sqrt{d}}{\frac{1}{r^2}[r^{d-1}\sin^{d-2}(\phi_1)\dots\sin(\phi_{d-2})]drd(\phi_1\otimes\dots\otimes\phi_{d-1})}}\\ &=\frac{(\pi\sqrt{d})^{d-2}}{d-2}\int_{[-\pi/2,\pi/2]^{d-1}}{[\sin^{d-2}(\phi_1)\dots\sin(\phi_{d-2})]d(\phi_1\otimes\dots\otimes\phi_{d-1})} \end{split} \end{equation} And this proves that in fact $\mid\theta\mid^{-2}$ is integrable on the compact set $[-\pi,\pi]^d$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .