Suppose that $f(x)$ is on continuously differentiable function, with bounded derivative, defined on $[0,1]$ such that $$ \lim\limits_{x\to1^-}f(x)-\lim\limits_{x\to0^+}f(x)=1 $$ The Fourier coefficients of f are defined by $$ \hat{f}(n)=\int_0^1f(x)e^{-2\pi inx}dx. $$ Prove that, as $n \to ±\infty$, we have: $$ \hat{f}(n)=\frac{i}{2\pi n}+o\left(\frac{1}{n}\right) $$
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Hint: An integration by parts shows that $$\hat f(n)=\frac1{-2\pi i n} +\frac1{2\pi in}\widehat{f'}(n).$$
David C. Ullrich
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I think the first part in your RHS is zero. – Gatsby Aug 12 '16 at 16:33
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@Gatsby I don't think so. Work it out again. The function $f$ doesn't have period $1$... – David C. Ullrich Aug 12 '16 at 16:44
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Yes, you are right. Thanks – Gatsby Aug 12 '16 at 17:08