Periodic continuous function which is integrable on $\mathbb{R}$

Let $f:\mathbb{R}\to\mathbb{R}$ be a $T$-periodic function, that is $f(t+T)=f(t)$ for all $t\in \mathbb{R}$. Assume that $$\int_0^{+\infty}|f(s)|ds<+\infty.$$ Now if we assume in addition that $f$ is continuous, my intuition tells me that we must have necessarily $f=0$, is this correct ?

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Yes, that is correct (for $T \neq 0$). –  Daniel Fischer Jun 27 '14 at 14:42

$$\int_0^{\infty} |f(s)| ds = \sum_{k = 0}^{\infty} \int_{kT}^{(k + 1)T} |f(s)| ds$$
This is correct. The way you can see this is by considering the maximum of $|f|$, call it $L$. For any $x$ such that $|f(x)|=L$, we have that $|f(y)| > \frac{L}{2}$ for all $|x-y| < \delta$ (for a sufficient choice of $\delta$). Can you see how to argue it from here?