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Let $f(x)=\frac{1}{2}-x$ on the interval $[0,1)$, and extend $f$ to be periodic in $\mathbb R$. Show that $\mathcal{F}(0)=0$, and $\mathcal{F}(k)=(2\pi ik)^{-1}$, if $k\ne0$, where my $\mathcal{F}(k) = \int\limits_{-\infty}^{\infty} f(x)\, e^{-2\pi i k x} \, \mathrm{d} x$.


I have proved that $\mathcal{F}(0)=0$, but for the other one, I am not able to as it needs two extra assumptions :

$(i)$ $k \in \mathbf Z$ $\And$

$(ii)$ The integral should be evaluated between just $0$ and $1$.

Can anyone explain me these two steps. Any type of help will be appreciated. Thanks in advance.

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  • $\begingroup$ The function $f$ is not integrable, are you sure that you did not mean the Fourier series? Are we talking distributions here? $\endgroup$ – copper.hat May 12 '17 at 4:18
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    $\begingroup$ yes i am sure that i need to evaluate the fourier transform. And it seems like the function is integrable, why did you said that it is not integrable $\endgroup$ – bunny May 12 '17 at 4:24
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    $\begingroup$ Because the function $f$ is not integrable on $\mathbb{R}$. $\endgroup$ – copper.hat May 12 '17 at 4:26
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    $\begingroup$ it is integrable and however the integral will come out to be 0 even $\endgroup$ – bunny May 12 '17 at 4:28
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    $\begingroup$ yes but you are taking the integral of $|f(x)|$ instead of $f(x)$ $\endgroup$ – bunny May 12 '17 at 4:33

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