Let $$ f(t) = \begin{cases} \cos(t) & \text{for } -8\pi \leq t \leq 8\pi\\ 0 & \text{otherwise} \end{cases} $$ Now $$ F(\omega) = \int_{-8 \pi}^{8 \pi} \cos(t)\, e^{-i \omega t} \,\mathrm dt =\frac{1}{2}\int_{-8 \pi}^{8 \pi} \left(e^{it}+e^{-it}\right)e^{i \omega t} \,\mathrm dt $$ by using Euler's formula.

Furthermore $$ \frac{1}{2}\int_{-8 \pi}^{8 \pi} e^{it(1-\omega)} + e^{-it(1+\omega)} \,\mathrm dt = \frac{1}{2}\left[\frac{e^{(1-\omega)it}}{(1-\omega)i}\right]_{-8\pi}^{8\pi} + \frac{1}{2}\left[\frac{-e^{-(1+\omega)it}}{(1+\omega)i}\right]_{-8\pi}^{8\pi} $$ Do i get a "nice result"? i guess the result should be able to be nicely written in terms of sin or something...

  • 1
    $\begingroup$ Why don't you at least finish the computation ? You can just use again the Euler formula ... $\endgroup$
    – LL 3.14
    Commented Nov 18, 2023 at 10:07

1 Answer 1


You are basically done. $\tfrac{1}{2}\left(\tfrac{1}{(1-w)i} e^{(1-w)it}\right)_{-8\pi}^{8\pi} = \tfrac{-i}{1-w}\tfrac{1}{2}(e^{-8\pi i w}-e^{8\pi i w})=\tfrac{i}{1-w}\tfrac{e^{8\pi i w}-e^{-8\pi i w}}{2}=\tfrac{\sin(8\pi w)}{w-1}$.

$\tfrac{1}{2}\left(\tfrac{-1}{(1+w)i}e^{-(1+w)it}\right)_{-8\pi}^{8\pi}=\tfrac{i}{(1+w)}\tfrac{1}{2}(e^{-(1+w)8\pi i}-e^{(1+w)8\pi i})=\tfrac{-i}{1+w}\tfrac{e^{8\pi i w}-e^{-8\pi i w}}{2}=\tfrac{\sin(8\pi w)}{w+1}$.

Combining, you get $F(w)=\tfrac{2w}{w^2-1} \sin(8\pi w)$, at least when $w\neq \pm 1$.

$F(\pm 1)=\int_{-8\pi}^{8\pi} \big(cos^2(t)\pm i\cos(t)\sin(t)\big)dt=8\int_{0}^{2\pi} \cos^2(t)dt$ since $\cos^2$ has period $2\pi$ and since $t\mapsto\cos(t)\sin(t)$ is an odd function that we are integrating over a symmetric domain.

The fact that $\int_0^{2\pi}\cos^2(t)dt = \pi$ is a standard exercise in integration by parts. Alternatively, one can check this by looking at the sum and the difference of the integrals $\int_0^{2\pi} \cos^2(t) dt$ and $\int_0^{2\pi} \sin^2(t)dt$.

This gives $F(\pm 1)=8\pi$. This can also be achieved as $\lim_{w\to\pm 1} \tfrac{2w}{w^2-1}\sin(8\pi w)$, which is not surprising in view of the Riemann-Lebesgue lemma since $f$ is integrable.


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