$\lim_{k\to\infty}\int_{1/k}^k f(x)\,dx \neq \int_0^{\infty}f(x)\,dx$

Question

I am trying to prove that $\lim_{k\to\infty}\int_{1/k}^k f(x)\,dx$ is not always equal to $\int_0^{\infty}f(x)\,dx$, where $\int_0^{\infty}f(x)\,dx=\lim_{R\to0}\lim_{S\to\infty}\int_R^Sf(x)\,dx$.

It seems that there should be a counterexample where $f(x)\to\infty$ as $x\to0$ and $x\to\infty$ with $\lim_{k\to\infty}\int_{1/k}^kf(x)\,dx=0$ but $\int_0^{\infty}f(x)\,dx$ does not exist.

For example, I tried $f(x)=|\ln x|$, but this does not work since then $\lim_{k\to\infty}\int_{1/k}^k f(x)\,dx$ does not exist.

For every example I have tried so far, they are the same. Can anyone suggest how to construct a counterexample?

Answer 1

Another example such that $$\lim_{A\to\infty}\int_{1/A}^Af(x)\,dx\neq\int_0^\infty f(x)\,dx.$$ Here $A$ tends to infinity through the real line rather than $\mathbb N$. Let $$f(x)=\begin{cases} \frac1x & x\geq1,\\ -\frac1x & 0<x<1. \end{cases}$$ So $\int_{1/A}^Af(x)\,dx=0$ for $A>1$ but $\int_0^\infty f(x)\,dx$ does not exist.

Answer 2

Let $f$ be any nice continuous function on $[0,1]$ with $f(0)=f(1)$ and $\int_0^{1} f(x) \, dx=0$ but $\int_0^{\frac 1 2} f(x) \, dx\neq 0$. Extend $f$ to a continuous periodic function with period $1$. Then the first limit is $0$ but the second limit does not exist

Answer 3

It's not a coincidence that $\int_0^\infty f$ does not exist in the two counterexamples given - it's easy to see that if $\int_0^\infty$ does exist (according to the definition you gave) then it does equal $\lim\int_{1/k}^kf$.