Counter-example to $\int_0^\infty f(x) dx=\lim_{t\to\infty} \int_{1/t}^t f(x) dx$

Question

I want to prove or disprove the statement that, for a function $f$ that is continuous on $(0,\infty)$, we have $\displaystyle{\int_0^\infty f(x)\ dx=\lim_{t\to\infty} \int_{1/t}^t f(x)\ dx}$.

My intuition is that it is false since it is similar to $\displaystyle{ \int_{-\infty}^\infty f(x)\ dx=\lim_{t\to\infty} \int_{-t}^t f(x)\ dx }$.

To prove it is false, I am looking for a function $f$ that satisfies

1. $\int_{1/t}^1 f(x)\ dx=-\int_{1}^t f(x)\ dx$ for any $t>1$
2. $f(1)=0$
3. and $\int_0^\infty f(x)\ dx$ is divergent.

It is too much for my purpose but I got inspired by the similar false statement above and its simple counter-example $f(x)=x$.

Using substitution $u=\frac{1}{x}$, I can show that if $f$ is solution of $f(\frac{1}{x})=x^2f(x)$, then the point $1$ is satisfied. Differentiating this functionial equation, we obtain that a differentiable solution is solution of $(1-x^8)y'-2(x^7-x^3)=0$, whose solutions are $y(x)=\frac{C}{\sqrt{1+x^4}}$. To satisfy the second point, we need $C=0$. But of course $\int_0^\infty\ 0\ dx$ is convergent.

I thought I would easily find a counter-example but I am stuck. Any idea?

Answer 1

What I was suggesting in the comment.

$$\int_{-\infty}^{\infty}2xdx=\int_{0}^{\infty}2\frac{\ln(y)}{y}dy$$ while $$\int_{1/t}^{t}2\frac{\ln(y)}{y}dy=(\ln(y))^2|_{1/t}^{t}=(\ln(t))^2-(\ln(1/t))^2=0$$