Series/integral of function

Question

I've tried to resolve an exercise and at some point I have to compare the integral : $ \int_{0}^{\infty} f(x) \, \mathrm{d}x $

with the series : $ \sum_{k=0}^{n-1} \int_{k\pi}^{(k+1)\pi} f(x) \, \mathrm{d}x $

in order to figure out if the integral converges or not.

In the exercise, we have $f(x) \geq 0$, and it's explained that as $f(x) \geq 0$ the series converges $\iff$ the integral converges.

But actually, I can't figure out why $f$ must have a constant sign, because maybe what I'm gonna written is absolutely wrong, but I want to say : $ \int_{0}^{n\pi} f(x) \, \mathrm{d}x = \sum_{k=0}^{n-1} \int_{k\pi}^{(k+1)\pi} f(x) \, \mathrm{d}x $, and so : $\lim\limits_{n \rightarrow +\infty} \int_{0}^{n\pi} f(x) \, \mathrm{d}x = \lim\limits_{n \rightarrow +\infty} \sum_{k=0}^{n-1} \int_{k\pi}^{(k+1)\pi} f(x) \, \mathrm{d}x $

whatever the sign of $f$, and then both the series and the integral have the same nature (convergent of divergent).

So, where is the mistake ?

Answer 1

Let $f(x)=\sin{2x}$. Then $\int_{k\pi}^{(k+1)\pi} \sin{2x} \, dx = 0$, but $\int_0^{\infty} \sin{2x} \, dx $ does not converge.

The point is that the partial sums only capture the behaviour of $\int_0^M f $ for certain values of $M$. It's the same as $\lim_{x \to \infty} \cos{x}$: if you just look at the values $2k\pi$, it looks like it converges to $1$, but we know this is not true.

You have (integral converges) $\implies $ (partial sums converge), but not the other direction.