I am confused about this thing

The definition of the Fourier transform of a function $f \in \mathbb{L}^1( \mathbb{R})$ (integrable function ): $$F(f)(x)=\int_{-\infty}^{\infty}f(x)e^{-i2\pi xt}dt$$ but if I want to compute the fourier transform of a periodic function ($f(x)=\sin(x)$ for example ) what i have to do 1) consider

$f(x)=\sin(x)$ for $x \in [0,2\pi]$

$ f(x)=0 $ else

2) compute the fourier trannformation of $\sin(x)$ as a function definied on $ \;\mathbb{R} (f(x)=\sin(x) $ for all $x \in \mathbb{R})$

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    $\begingroup$ The Fourier transform does not exist as an ordinary function. The integral does not converge. $\endgroup$ – Kavi Rama Murthy May 21 at 23:36
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    $\begingroup$ Let $\hat{f}_n(x) = \int_{-n}^n f(x)e^{-2i \pi xt}dt$ when $f$ is $L^1_{loc}$ and $T$-periodic then $\lim_{n \to \infty}\hat{f}_n$ converges only in the sense of distributions, to $\hat{f}(x)=\sum_k c_k(f) \delta(x-k/T)$ where $c_k(f) = \frac{1}{T} \int_0^T f(t)e^{-2i \pi kt/T}dt$ is the Fourier series coefficient. The inverse Fourier transform is just the Fourier series, which converges in the sense of distributions. $\endgroup$ – reuns May 22 at 0:44

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