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How does one show that $\chi_{[1, \infty)}1/x$ is not (Lebesgue) integrable?


What I could think of is as follows:

Letting $f(x)=1/x$ (defined for $x\geq 1$), define $$ f_n(x)=f\chi_{[1, n)}(x). $$

Each $f_n$ is, therefore, Riemann integrable on $[1, n)$ with value $\ln n$, hence integrable there. As $0\leq f_n\nearrow f$ on $[1, \infty)$, the monotone increasing theorem says $$ \int_{[1, \infty)}f_n\nearrow\int_{[1, \infty)} f $$ and so $\int_{[1, \infty)} f=\infty$ since $\ln n\nearrow\infty$.


Is there a more obvious reason why the given integral isn't finite? It seems that my method needs quite some modification if we go to $n$-dimensional integrals of $$ f(x)=\frac{1}{|x|}\chi_{|x|>1}. $$

  • 1
    Hint: switch to polar coordinates2012-10-26

4 Answers 4