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I have to prove that this integral is convergent:

$ int_vertxivert geq 1 vertxivert^-2[fracn2]-2 dxi $

where $ n $ is the dimension of $ R^n $ and $ [ cdot] $ denotes the floor function. Maybe polar coordinates can be used, but i don't know how to prove that is convergent. Can anyone help me?

Thank you.

In order to define polar coordinates in $n$ dimensions, we need $n+1$ angular variables. One way to set this up is to take a hint from three-dimensional spherical coordinates; define $x_1=rcostheta_1$, $x_2=rsintheta_1costheta_2$, $x_3=rsintheta_1sintheta_2costheta_3$, and so on until we close it with $x_n=rsintheta_1sintheta_2sintheta_3cdots sintheta_n-1$. In transforming to these coordinates, the Jacobian will be something involving the angular variables times $r^n-1$. If we integrate a radially symmetric function $f(|x|)$ over the radially symmetric region $|x|>1$, we get a constant from the $theta$ integrals and
$$int_xf(x),dx_1,dx_2,cdots,dx_n = A(n)int_1^inftyr^n-1f(r),dr$$
where $A(n)$ depends only on $n$. For $f(r)=r^-2lfloor n/2rfloor-2$, that leads us to the integral of $r^-3$ or $r^-2$, depending on whether $n$ is even or odd. You should know how to handle the convergence of a one-dimensional integral like this.

We can also avoid having to define these coordinates. It is well-known that every norm on $mathbbR^n$ is equivalent - so let's work with a more convenient norm. The $infty$-norm $|x|_infty=maxx_1,x_2,dots,x_n$ is particularly easy to work with for integrals, as it will naturally depend (locally) on only one of the variables. In converting between norms, we have that $|x|_2le sqrtn|x|_infty$ and $|x|_infty le |x|_2$.

So then, increasing first the region and then the integrand,
$$int_x|x|_2^-2lfloor n/2rfloor-2,dx le int_|x|_2^-2lfloor n/2rfloor-2,dx le int_|x|_infty^-2lfloor n/2rfloor-2,dx=I$$
Now, we integrate that. Split the region based on which coordinate is largest and whether it's positive or negative; we get $2n$ regions this way, of which the first is $x_1ge |x_2|,|x_3|,dots,|x_n|$. By symmetry, the integral over each of these regions is equal, and
$$I = 2nint_1/sqrtn^inftyint_-x_1^x_1cdots int_-x_1^x_1x_1^-2lfloor n/2rfloor-2,dx_n,cdots,dx_2,dx_1 = 2nint_1/sqrtn^infty2^n-1x_1^n-2lfloor n/2rfloor-3,dx_1$$
Then $n-2lfloor n/2rfloor=begincases0&ntext is even\1&ntext is oddendcases$. For even $n$, we get
$$int_x|x|_2^-2lfloor n/2rfloor-2,dx le I=2^n nint_1/sqrtn^inftyx_1^-3,dx=left. -2^n-1 nx_1^-2right|_1/sqrtn^infty=2^n-1$$
For odd $n$, we instead have
$$int_x|x|_2^-2lfloor n/2rfloor-2,dx le I=2^n nint_1/sqrtn^inftyx_1^-2,dx=left. -2^n nx_1^-1right|_1/sqrtn^infty=2^nsqrtn$$
Either way, the original integral converges by comparison.

Just you polar coordinates and consider the cases $n$ even and $n$ odd separately. The integral is convergent in both cases.