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Let $f\in L^1(\mathbb{T})$ where $\mathbb{T}$ is the unit circle in the complex plane. I need to calculate $\frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\int_{0}^{t}f(e^{i \tau})d\tau\right) e^{(-int)}dt$. The answer is $\frac{\hat{f}(n)}{in}$. Is my following approach correct?

$\frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\int_{0}^{t}f(e^{i \tau})d\tau\right) e^{(-int)}dt$

=$\frac{1}{2\pi}\int_{-\pi}^{\pi}f(e^{it})\left(\int_{0}^{t} e^{(-in \tau)}d\tau\right)dt$ (if we substitute $\tau=t$)

=$\frac{\hat{f}(n)}{in}$

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  • $\begingroup$ No, your argument is not correct. You have to integrate by parts to get the formula. $\endgroup$ Aug 23, 2018 at 11:45
  • $\begingroup$ @KaviRamaMurthy OK with integration by parts I get $\left[\int_{0}^{t}f(e^{i \tau})d\tau\; e^{(-int)}\right]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}f(e^{it})\int e^{(int)}dt$. Why is the first term zero? $\endgroup$
    – user31459
    Aug 23, 2018 at 12:14
  • $\begingroup$ The problem is $\int_0^{t} f(e^{is}) \, ds$ need not be a periodic function. If it is periodic you will get the first term as $0$. Otherwise you modify the function to make it periodic. $\endgroup$ Aug 23, 2018 at 23:21

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