# Fourier transform of $\mathrm{cos}(t)$

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...

• Why don't you at least finish the computation ? You can just use again the Euler formula ... Commented Nov 18, 2023 at 10:07

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.