Example of a continuous function $f:(0,1) \rightarrow \mathbb{R}$ such that $\int_{1/n}^1 f \, dx = 0$ for all $n$ but $\int_0^1 f \, dx$ DNE?

Question

I'm looking for an example of a continuous function $f:(0,1) \rightarrow \mathbb{R}$ such that $\int_{1/n}^1 f \, dx = 0$ for all $n$ but $\int_0^1 f \, dx$ does not exist.

I know we can consider the function $F(x) = \int_x^1 f(x) \, dx$. So the above would be equivalent to finding $F(x)$ s.t. $F(1/n) = 0$ for all $n$, but $\lim_{x \rightarrow 0} F(x)$ does not exist. Then taking the derivative of $F$ we will get $f$.

Thanks...

Answer 1

Let $F(x)$ be defined as $F(x)=\sin(\pi /x)$ for $x\in (0,1)$. Clearly, $F(x)$ is differentiable for $x\in(0,1)$ with

$$f(x)\equiv -F'(x) =\frac{\pi \cos(\pi/x)}{x^2}$$

Note that $F'(x)$ is continuous on $(0,1)$.

We have, therefore, that

$$\int_{1/n}^1 \frac{\pi \cos(\pi/x)}{x^2}\,dx=-\left.\sin(\pi/x)\right|_{x=1/n}^1=0$$

for all $n\in \mathbb{N}$.

Moreover, the integral $\int_0^1 \frac{\pi \cos(\pi/x)}{x^2}\,dx$ fails to exist.

Finally, $f(x)=\frac{\pi \cos(\pi/x)}{x^2}\in C(0,1)$ such that

$$\int_{1/n}^1 f(x)\,dx=0$$

for all $n\in \mathbb{N}$ and such that $\int_0^1 f(x)\,dx$ fails to exist.