Suppose we have $2\pi$ periodic function $f:\mathbb{R} \rightarrow \mathbb{C}$ that is integrable on $[0,2\pi]$ where it holds that $\int^{2\pi}_0 f(x) dx = 0$ Define this function $$h:\mathbb{R} \rightarrow \mathbb{C}:h(t) = \int^t_0 f(x) dx. $$ How can we see that $h$ is well defined? Usually it seems quite obvious how to prove "well-definedness" but I'm a bit confused here. Any help is appreciated!
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1perhaps the author is asking if it is well defined because f is integrable only on $[0,2\pi]$, but using the fact that it is periodic, it must be integrable on any interval. – Ilovemath Jan 23 '22 at 13:55
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@Ilovemath hmm I'm still not entirely convinced – Geigercounter Jan 23 '22 at 13:56
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add a answer! I hope it helps. – Ilovemath Jan 23 '22 at 14:14
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for example, write $t=t_0+2k\pi$, with $t_0 \in [0,2\pi]$ then
$$h(t) = \int_0^{t_0+2k\pi}f(x)dx = \int_0^{t_0}f(x)dx+\int_{t_0}^{t_0+2\pi}f(x)dx+ \cdots+\int_{t_0+2(k-1)\pi}^{[t_0+2(k-1)\pi]+2\pi}f(x)dx$$ successively use this result enter link description here.
With this result, all that matters is the fact that $\int_0^{t_0}f(x)dx$ exists.
Ilovemath
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so this expression for $h(t)$ just reduces to the first term on the right? – Geigercounter Jan 23 '22 at 14:18
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