It's a well-known fact in Fourier analysis that the sawtooth function has a convergent (pointwise) Fourier series at all points, including at the discontinuities. Specifically, if we define the sawtooth function as the 2-periodic function $f(x) = x$ for $x \in [-1, 1]$, we can show the Fourier series converges to $$ f(x) = \frac{2}{\pi} \sum_{n \geq 1} \frac{(-1)^{n + 1}sin(n \pi x)}{n} $$

For $x \in (-1, 1)$, and $\hat{f}(x) = 0$, $x \in \{-1, 1\}$.

If, however, we approach $x = 1$ from the left, I've been told the partial sums $$ S^{f}_{n}(x) = \frac{1}{2} \int_{-1}^{1} f(x - t) \frac{sin\big((n + \frac{1}{2})t\big)}{sin\big(\frac{t}{2}\big)}dt $$

Have the property that

$$ S^{f}_{n}(1 - \frac{1}{n}) \rightarrow \frac{2}{\pi} \int_{0}^{1} \frac{sin(\pi t)}{t}dt $$

I'm not sure how to show this. Approaching from the standard definition of the Fourier coefficients seems to get somewhere close, but I'm not entirely sure how to evaluate the resulting integral. I've also tried integrating with the Dirichlet kernel as shown above, but this seems to be even further from the desired result.



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