Does $\int_{\mathbb R^n} (1 + |x|^2)^{-1} dx$ converge in the Riemann sense?

Question

Recently I was reading an article and came across the following integral:

$$\int_{\mathbb R^n}\dfrac{1}{1+|x|^2}\ dx$$

Is this integral convergent in the Riemann sense?

Answer 1

The integrand is positive and smooth so Riemann or Lebesgue makes no difference. Change coordinates as follows. $$\int_{\mathbb{R}^n} {dx\over 1 + |x|^2} = \int_0^\infty\int_{S^{n-1}} {r^{n-1}\,d\sigma\,dr\over 1 + r^2} = c_n \int_0^\infty {r^{n-1}\,dr\over 1 + r^2}$$ This converges if $n = 1$. If $n\ge 2$, it diverges.

Answer 2

It converges (in the improper Riemann sense) if and only if $n < 2$, i.e. if $n=1$. This follows immediately by comparison from the following theorem.

Suppose $E \subseteq \mathbb{R}^n$ and $f:E\to \mathbb{R}$ is continuous. Let $x_0 \in \mathbb{R}^n\setminus E$ and assume $a, C >0$ are such that $|f|$ is bounded by $$ \frac{C}{|x-x_0|^a} $$ for all $x \in E$. Then

(1) If $E$ is Peano-Jordan measurable and $a < n$, then $f$ is Riemann integrable in the improper sense with finite integral.

(2) If $E \subseteq \mathbb{R}^n \setminus B(x_0,r)$ admits an exhausting sequence and $a >n$, then $f$ is Riemann integrable in the improper sense with finite integral.